Endpoint Extrema — Helicopter Method — Dynamic Practice
Endpoint Extrema — Helicopter Method — Dynamic Practice. Practice questions to deepen understanding of extrema at the edges of the domain using the helicopter method. Interactive math practice with instant feedback.
Dynamic practice in endpoint extrema using the "helicopter" method — visualizing the function value just before/after the domain edge to identify endpoint extrema. New questions every attempt.
Question 1
2.38 pts
Check if there is an extremum [0, 5] at the Left edge \(x = 0\):
\(f(x) = x^{2}+2x+2\)
\(f(x) = x^{2}+2x+2\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = x^{2}+2x+2\)
The function: \(f(x) = x^{2}+2x+2\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 2x+2\)
🎯 Step 2: Substitute the edge point Left
\(x = 0\)
\(f'(0) = 2\)
🔍 Step 3: Check the sign
\(f'(0) = 2\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising from the left edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = 0\)
Question 2
2.38 pts
Check if there is an extremum [-4, 0] at the Right edge \(x = 0\):
\(f(x) = x^{2}-2x\)
\(f(x) = x^{2}-2x\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = x^{2}-2x\)
The function: \(f(x) = x^{2}-2x\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 2x-2\)
🎯 Step 2: Substitute the edge point Right
\(x = 0\)
\(f'(0) = -2\)
🔍 Step 3: Check the sign
\(f'(0) = -2\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending to the right edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = 0\)
Question 3
2.38 pts
Check if there is an extremum [-7, -3] at the Right edge \(x = -3\):
\(f(x) = 3x^{2}+2x+2\)
\(f(x) = 3x^{2}+2x+2\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 3x^{2}+2x+2\)
The function: \(f(x) = 3x^{2}+2x+2\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 6x+2\)
🎯 Step 2: Substitute the edge point Right
\(x = -3\)
\(f'(-3) = -16\)
🔍 Step 3: Check the sign
\(f'(-3) = -16\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending to the right edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = -3\)
Question 4
2.38 pts
Check if there is an extremum [1, 5] at the Left edge \(x = 1\):
\(f(x) = x^{2}-2x-1\)
\(f(x) = x^{2}-2x-1\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = x^{2}-2x-1\)
The function: \(f(x) = x^{2}-2x-1\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 2x-2\)
🎯 Step 2: Substitute the edge point Left
\(x = 1\)
\(f'(1) = 0\)
🔍 Step 3: Check the sign
\(f'(1) = 0\) = \(0\)
The function Horizontal ─
The function Horizontal ─
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Horizontal at the edge → no extremum at the edge
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: No extremum at the edge
Question 5
2.38 pts
Check if there is an extremum [3, 8] at the Left edge \(x = 3\):
\(f(x) = 2x^{2}-3x\)
\(f(x) = 2x^{2}-3x\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 2x^{2}-3x\)
The function: \(f(x) = 2x^{2}-3x\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 4x-3\)
🎯 Step 2: Substitute the edge point Left
\(x = 3\)
\(f'(3) = 9\)
🔍 Step 3: Check the sign
\(f'(3) = 9\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising from the left edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = 3\)
Question 6
2.38 pts
Check if there is an extremum [-8, -3] at the Right edge \(x = -3\):
\(f(x) = x^{2}+2x\)
\(f(x) = x^{2}+2x\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = x^{2}+2x\)
The function: \(f(x) = x^{2}+2x\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 2x+2\)
🎯 Step 2: Substitute the edge point Right
\(x = -3\)
\(f'(-3) = -4\)
🔍 Step 3: Check the sign
\(f'(-3) = -4\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending to the right edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = -3\)
Question 7
2.38 pts
Check if there is an extremum [3, 9] at the Left edge \(x = 3\):
\(f(x) = 3x^{2}-x-2\)
\(f(x) = 3x^{2}-x-2\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 3x^{2}-x-2\)
The function: \(f(x) = 3x^{2}-x-2\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 6x-1\)
🎯 Step 2: Substitute the edge point Left
\(x = 3\)
\(f'(3) = 17\)
🔍 Step 3: Check the sign
\(f'(3) = 17\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising from the left edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = 3\)
Question 8
2.38 pts
Check if there is an extremum [-3, 2] at the Right edge \(x = 2\):
\(f(x) = 3x^{2}+2x+3\)
\(f(x) = 3x^{2}+2x+3\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 3x^{2}+2x+3\)
The function: \(f(x) = 3x^{2}+2x+3\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 6x+2\)
🎯 Step 2: Substitute the edge point Right
\(x = 2\)
\(f'(2) = 14\)
🔍 Step 3: Check the sign
\(f'(2) = 14\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising to the right edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = 2\)
Question 9
2.38 pts
Check if there is an extremum [-4, 1] at the Right edge \(x = 1\):
\(f(x) = 2x^{2}-x-3\)
\(f(x) = 2x^{2}-x-3\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 2x^{2}-x-3\)
The function: \(f(x) = 2x^{2}-x-3\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 4x-1\)
🎯 Step 2: Substitute the edge point Right
\(x = 1\)
\(f'(1) = 3\)
🔍 Step 3: Check the sign
\(f'(1) = 3\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising to the right edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = 1\)
Question 10
2.38 pts
Check if there is an extremum [-3, 3] at the Left edge \(x = -3\):
\(f(x) = 3x^{2}+2x\)
\(f(x) = 3x^{2}+2x\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 3x^{2}+2x\)
The function: \(f(x) = 3x^{2}+2x\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 6x+2\)
🎯 Step 2: Substitute the edge point Left
\(x = -3\)
\(f'(-3) = -16\)
🔍 Step 3: Check the sign
\(f'(-3) = -16\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending from the left edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = -3\)
Question 11
2.38 pts
Check if there is an extremum [3, 8] at the Left edge \(x = 3\):
\(f(x) = 3x^{2}+2x-2\)
\(f(x) = 3x^{2}+2x-2\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 3x^{2}+2x-2\)
The function: \(f(x) = 3x^{2}+2x-2\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 6x+2\)
🎯 Step 2: Substitute the edge point Left
\(x = 3\)
\(f'(3) = 20\)
🔍 Step 3: Check the sign
\(f'(3) = 20\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising from the left edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = 3\)
Question 12
2.38 pts
Check if there is an extremum [-4, 2] at the Right edge \(x = 2\):
\(f(x) = 3x^{2}+2x+3\)
\(f(x) = 3x^{2}+2x+3\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 3x^{2}+2x+3\)
The function: \(f(x) = 3x^{2}+2x+3\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 6x+2\)
🎯 Step 2: Substitute the edge point Right
\(x = 2\)
\(f'(2) = 14\)
🔍 Step 3: Check the sign
\(f'(2) = 14\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising to the right edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = 2\)
Question 13
2.38 pts
Check if there is an extremum [-5, 0] at the Right edge \(x = 0\):
\(f(x) = 2x^{2}+2x-2\)
\(f(x) = 2x^{2}+2x-2\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 2x^{2}+2x-2\)
The function: \(f(x) = 2x^{2}+2x-2\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 4x+2\)
🎯 Step 2: Substitute the edge point Right
\(x = 0\)
\(f'(0) = 2\)
🔍 Step 3: Check the sign
\(f'(0) = 2\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising to the right edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = 0\)
Question 14
2.38 pts
Check if there is an extremum [0, 3] at the Right edge \(x = 3\):
\(f(x) = x^{2}+x+1\)
\(f(x) = x^{2}+x+1\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = x^{2}+x+1\)
The function: \(f(x) = x^{2}+x+1\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 2x+1\)
🎯 Step 2: Substitute the edge point Right
\(x = 3\)
\(f'(3) = 7\)
🔍 Step 3: Check the sign
\(f'(3) = 7\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising to the right edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = 3\)
Question 15
2.38 pts
Check if there is an extremum [-2, 4] at the Left edge \(x = -2\):
\(f(x) = 3x^{2}+3x+2\)
\(f(x) = 3x^{2}+3x+2\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 3x^{2}+3x+2\)
The function: \(f(x) = 3x^{2}+3x+2\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 6x+3\)
🎯 Step 2: Substitute the edge point Left
\(x = -2\)
\(f'(-2) = -9\)
🔍 Step 3: Check the sign
\(f'(-2) = -9\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending from the left edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = -2\)
Question 16
2.38 pts
Check if there is an extremum [-4, 2] at the Right edge \(x = 2\):
\(f(x) = 4x^{2}-5\)
\(f(x) = 4x^{2}-5\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 4x^{2}-5\)
The function: \(f(x) = 4x^{2}-5\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 8x\)
🎯 Step 2: Substitute the edge point Right
\(x = 2\)
\(f'(2) = 16\)
🔍 Step 3: Check the sign
\(f'(2) = 16\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising to the right edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = 2\)
Question 17
2.38 pts
Check if there is an extremum [-7, -1] at the Right edge \(x = -1\):
\(f(x) = 4x^{2}-2\)
\(f(x) = 4x^{2}-2\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 4x^{2}-2\)
The function: \(f(x) = 4x^{2}-2\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 8x\)
🎯 Step 2: Substitute the edge point Right
\(x = -1\)
\(f'(-1) = -8\)
🔍 Step 3: Check the sign
\(f'(-1) = -8\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending to the right edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = -1\)
Question 18
2.38 pts
Check if there is an extremum [-6, -3] at the Right edge \(x = -3\):
\(f(x) = x^{2}-3x+4\)
\(f(x) = x^{2}-3x+4\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = x^{2}-3x+4\)
The function: \(f(x) = x^{2}-3x+4\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 2x-3\)
🎯 Step 2: Substitute the edge point Right
\(x = -3\)
\(f'(-3) = -9\)
🔍 Step 3: Check the sign
\(f'(-3) = -9\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending to the right edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = -3\)
Question 19
2.38 pts
Check if there is an extremum [-3, 2] at the Left edge \(x = -3\):
\(f(x) = 5x^{2}+2x\)
\(f(x) = 5x^{2}+2x\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 5x^{2}+2x\)
The function: \(f(x) = 5x^{2}+2x\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 10x+2\)
🎯 Step 2: Substitute the edge point Left
\(x = -3\)
\(f'(-3) = -28\)
🔍 Step 3: Check the sign
\(f'(-3) = -28\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending from the left edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = -3\)
Question 20
2.38 pts
Check if there is an extremum [-4, 1] at the Right edge \(x = 1\):
\(f(x) = x^{2}-5x-4\)
\(f(x) = x^{2}-5x-4\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = x^{2}-5x-4\)
The function: \(f(x) = x^{2}-5x-4\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 2x-5\)
🎯 Step 2: Substitute the edge point Right
\(x = 1\)
\(f'(1) = -3\)
🔍 Step 3: Check the sign
\(f'(1) = -3\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending to the right edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = 1\)
Question 21
2.38 pts
Check if there is an extremum [-1, 2] at the Right edge \(x = 2\):
\(f(x) = 3x^{2}-x+3\)
\(f(x) = 3x^{2}-x+3\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 3x^{2}-x+3\)
The function: \(f(x) = 3x^{2}-x+3\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 6x-1\)
🎯 Step 2: Substitute the edge point Right
\(x = 2\)
\(f'(2) = 11\)
🔍 Step 3: Check the sign
\(f'(2) = 11\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising to the right edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = 2\)
Question 22
2.38 pts
Check if there is an extremum [-1, 2] at the Right edge \(x = 2\):
\(f(x) = 3x^{2}-4x-4\)
\(f(x) = 3x^{2}-4x-4\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 3x^{2}-4x-4\)
The function: \(f(x) = 3x^{2}-4x-4\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 6x-4\)
🎯 Step 2: Substitute the edge point Right
\(x = 2\)
\(f'(2) = 8\)
🔍 Step 3: Check the sign
\(f'(2) = 8\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising to the right edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = 2\)
Question 23
2.38 pts
Check if there is an extremum [-2, 3] at the Right edge \(x = 3\):
\(f(x) = 5x^{2}+2\)
\(f(x) = 5x^{2}+2\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 5x^{2}+2\)
The function: \(f(x) = 5x^{2}+2\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 10x\)
🎯 Step 2: Substitute the edge point Right
\(x = 3\)
\(f'(3) = 30\)
🔍 Step 3: Check the sign
\(f'(3) = 30\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising to the right edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = 3\)
Question 24
2.38 pts
Check if there is an extremum [3, 9] at the Left edge \(x = 3\):
\(f(x) = x^{2}+2\)
\(f(x) = x^{2}+2\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = x^{2}+2\)
The function: \(f(x) = x^{2}+2\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 2x\)
🎯 Step 2: Substitute the edge point Left
\(x = 3\)
\(f'(3) = 6\)
🔍 Step 3: Check the sign
\(f'(3) = 6\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising from the left edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = 3\)
Question 25
2.38 pts
Check if there is an extremum [-2, 2] at the Right edge \(x = 2\):
\(f(x) = x^{2}-3x-5\)
\(f(x) = x^{2}-3x-5\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = x^{2}-3x-5\)
The function: \(f(x) = x^{2}-3x-5\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 2x-3\)
🎯 Step 2: Substitute the edge point Right
\(x = 2\)
\(f'(2) = 1\)
🔍 Step 3: Check the sign
\(f'(2) = 1\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising to the right edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = 2\)
Question 26
2.38 pts
Check if there is an extremum [3, 6] at the Left edge \(x = 3\):
\(f(x) = 4x^{2}-4x-2\)
\(f(x) = 4x^{2}-4x-2\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 4x^{2}-4x-2\)
The function: \(f(x) = 4x^{2}-4x-2\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 8x-4\)
🎯 Step 2: Substitute the edge point Left
\(x = 3\)
\(f'(3) = 20\)
🔍 Step 3: Check the sign
\(f'(3) = 20\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising from the left edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = 3\)
Question 27
2.38 pts
Check if there is an extremum [-1, 5] at the Left edge \(x = -1\):
\(f(x) = x^{2}+2x-1\)
\(f(x) = x^{2}+2x-1\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = x^{2}+2x-1\)
The function: \(f(x) = x^{2}+2x-1\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 2x+2\)
🎯 Step 2: Substitute the edge point Left
\(x = -1\)
\(f'(-1) = 0\)
🔍 Step 3: Check the sign
\(f'(-1) = 0\) = \(0\)
The function Horizontal ─
The function Horizontal ─
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Horizontal at the edge → no extremum at the edge
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: No extremum at the edge
Question 28
2.38 pts
Check if there is an extremum [-8, -2] at the Right edge \(x = -2\):
\(f(x) = 2x^{2}+5x\)
\(f(x) = 2x^{2}+5x\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 2x^{2}+5x\)
The function: \(f(x) = 2x^{2}+5x\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 4x+5\)
🎯 Step 2: Substitute the edge point Right
\(x = -2\)
\(f'(-2) = -3\)
🔍 Step 3: Check the sign
\(f'(-2) = -3\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending to the right edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = -2\)
Question 29
2.38 pts
Check if there is an extremum [-6, -3] at the Right edge \(x = -3\):
\(f(x) = 5x^{2}-6x+7\)
\(f(x) = 5x^{2}-6x+7\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 5x^{2}-6x+7\)
The function: \(f(x) = 5x^{2}-6x+7\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 10x-6\)
🎯 Step 2: Substitute the edge point Right
\(x = -3\)
\(f'(-3) = -36\)
🔍 Step 3: Check the sign
\(f'(-3) = -36\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending to the right edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = -3\)
Question 30
2.38 pts
Check if there is an extremum [-5, 0] at the Right edge \(x = 0\):
\(f(x) = 3x^{2}-3x+3\)
\(f(x) = 3x^{2}-3x+3\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 3x^{2}-3x+3\)
The function: \(f(x) = 3x^{2}-3x+3\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 6x-3\)
🎯 Step 2: Substitute the edge point Right
\(x = 0\)
\(f'(0) = -3\)
🔍 Step 3: Check the sign
\(f'(0) = -3\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending to the right edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = 0\)
Question 31
2.38 pts
Check if there is an extremum [-8, -3] at the Right edge \(x = -3\):
\(f(x) = 5x^{2}+3x-3\)
\(f(x) = 5x^{2}+3x-3\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 5x^{2}+3x-3\)
The function: \(f(x) = 5x^{2}+3x-3\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 10x+3\)
🎯 Step 2: Substitute the edge point Right
\(x = -3\)
\(f'(-3) = -27\)
🔍 Step 3: Check the sign
\(f'(-3) = -27\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending to the right edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = -3\)
Question 32
2.38 pts
Check if there is an extremum [-5, -1] at the Right edge \(x = -1\):
\(f(x) = 8x^{2}+x+6\)
\(f(x) = 8x^{2}+x+6\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 8x^{2}+x+6\)
The function: \(f(x) = 8x^{2}+x+6\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 16x+1\)
🎯 Step 2: Substitute the edge point Right
\(x = -1\)
\(f'(-1) = -15\)
🔍 Step 3: Check the sign
\(f'(-1) = -15\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending to the right edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = -1\)
Question 33
2.38 pts
Check if there is an extremum [-4, -1] at the Right edge \(x = -1\):
\(f(x) = 5x^{2}-6x\)
\(f(x) = 5x^{2}-6x\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 5x^{2}-6x\)
The function: \(f(x) = 5x^{2}-6x\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 10x-6\)
🎯 Step 2: Substitute the edge point Right
\(x = -1\)
\(f'(-1) = -16\)
🔍 Step 3: Check the sign
\(f'(-1) = -16\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending to the right edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = -1\)
Question 34
2.38 pts
Check if there is an extremum [-1, 5] at the Left edge \(x = -1\):
\(f(x) = 2x^{2}+4\)
\(f(x) = 2x^{2}+4\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 2x^{2}+4\)
The function: \(f(x) = 2x^{2}+4\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 4x\)
🎯 Step 2: Substitute the edge point Left
\(x = -1\)
\(f'(-1) = -4\)
🔍 Step 3: Check the sign
\(f'(-1) = -4\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending from the left edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = -1\)
Question 35
2.38 pts
Check if there is an extremum [0, 6] at the Left edge \(x = 0\):
\(f(x) = 4x^{2}-x+3\)
\(f(x) = 4x^{2}-x+3\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 4x^{2}-x+3\)
The function: \(f(x) = 4x^{2}-x+3\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 8x-1\)
🎯 Step 2: Substitute the edge point Left
\(x = 0\)
\(f'(0) = -1\)
🔍 Step 3: Check the sign
\(f'(0) = -1\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending from the left edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = 0\)
Question 36
2.38 pts
Check if there is an extremum [-5, 0] at the Right edge \(x = 0\):
\(f(x) = 8x^{2}-2x-1\)
\(f(x) = 8x^{2}-2x-1\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 8x^{2}-2x-1\)
The function: \(f(x) = 8x^{2}-2x-1\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Right
🔍 Step 3: Check the sign
\(f'(x) = 16x-2\)
🎯 Step 2: Substitute the edge point Right
\(x = 0\)
\(f'(0) = -2\)
🔍 Step 3: Check the sign
\(f'(0) = -2\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending to the right edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = 0\)
Question 37
2.38 pts
Check if there is an extremum [-1, 4] at the Left edge \(x = -1\):
\(f(x) = 8x^{2}+2x-5\)
\(f(x) = 8x^{2}+2x-5\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 8x^{2}+2x-5\)
The function: \(f(x) = 8x^{2}+2x-5\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 16x+2\)
🎯 Step 2: Substitute the edge point Left
\(x = -1\)
\(f'(-1) = -14\)
🔍 Step 3: Check the sign
\(f'(-1) = -14\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending from the left edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = -1\)
Question 38
2.38 pts
Check if there is an extremum [2, 6] at the Left edge \(x = 2\):
\(f(x) = 6x^{2}-5x+1\)
\(f(x) = 6x^{2}-5x+1\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 6x^{2}-5x+1\)
The function: \(f(x) = 6x^{2}-5x+1\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 12x-5\)
🎯 Step 2: Substitute the edge point Left
\(x = 2\)
\(f'(2) = 19\)
🔍 Step 3: Check the sign
\(f'(2) = 19\) > \(0\)
The function increasing ↗
The function increasing ↗
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Rising from the left edge → this is a Minimum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Minimum at the edge \(x = 2\)
Question 39
2.38 pts
Check if there is an extremum [-1, 5] at the Left edge \(x = -1\):
\(f(x) = 5x^{2}-2x+6\)
\(f(x) = 5x^{2}-2x+6\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 5x^{2}-2x+6\)
The function: \(f(x) = 5x^{2}-2x+6\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 10x-2\)
🎯 Step 2: Substitute the edge point Left
\(x = -1\)
\(f'(-1) = -12\)
🔍 Step 3: Check the sign
\(f'(-1) = -12\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending from the left edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = -1\)
Question 40
2.38 pts
Check if there is an extremum [-3, 1] at the Left edge \(x = -3\):
\(f(x) = 4x^{2}-2\)
\(f(x) = 4x^{2}-2\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 4x^{2}-2\)
The function: \(f(x) = 4x^{2}-2\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 8x\)
🎯 Step 2: Substitute the edge point Left
\(x = -3\)
\(f'(-3) = -24\)
🔍 Step 3: Check the sign
\(f'(-3) = -24\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending from the left edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = -3\)
Question 41
2.38 pts
Check if there is an extremum [0, 5] at the Left edge \(x = 0\):
\(f(x) = 8x^{2}-x+5\)
\(f(x) = 8x^{2}-x+5\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 8x^{2}-x+5\)
The function: \(f(x) = 8x^{2}-x+5\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 16x-1\)
🎯 Step 2: Substitute the edge point Left
\(x = 0\)
\(f'(0) = -1\)
🔍 Step 3: Check the sign
\(f'(0) = -1\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending from the left edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = 0\)
Question 42
2.38 pts
Check if there is an extremum [0, 3] at the Left edge \(x = 0\):
\(f(x) = 2x^{2}-3x\)
\(f(x) = 2x^{2}-3x\)
Explanation:
Solution - Extrema at edge (helicopter method 🚁):
The function: \(f(x) = 2x^{2}-3x\)
The function: \(f(x) = 2x^{2}-3x\)
🔢 Step 1: Calculate the derivative
🎯 Step 2: Substitute the edge point Left
🔍 Step 3: Check the sign
\(f'(x) = 4x-3\)
🎯 Step 2: Substitute the edge point Left
\(x = 0\)
\(f'(0) = -3\)
🔍 Step 3: Check the sign
\(f'(0) = -3\) < \(0\)
The function decreasing ↘
The function decreasing ↘
🚁 " + BLL.LocalizationService.T("generator.extrema.method.at_edge_short", lang) + @":
🚁 Descending from the left edge → this is a Maximum at point!
📝 " + BLL.LocalizationService.T("generator.angles.rule_label", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
" + BLL.LocalizationService.T("generator.extrema.heli_rule_table", lang) + @"
Answer: Maximum at the edge \(x = 0\)