AP Calculus BC 4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms - FRQs - Exam Style Questions
Calc-Ok Question
The function \(f\) is defined on the closed interval \([-2,8]\) and satisfies \(f(2)=1\). The graph of \(f’\) (two line segments and a semicircle) is shown.
(a) Does \(f\) have a relative minimum, a relative maximum, or neither at \(x=6\)? Give a reason.
(b) On what open interval(s), if any, is the graph of \(f\) concave down? Give a reason.
(c) Find the value of \(\displaystyle \lim_{x\to 2}\frac{6f(x)-3x}{x^{2}-5x+6}\), or show that it does not exist. Justify your answer.
(d) Find the absolute minimum value of \(f\) on \([-2,8]\). Justify your answer.
Most-appropriate topic codes (CED):
• TOPIC 5.4: Using the First Derivative Test to Determine Relative (Local) Extrema — part (a)
• TOPIC 5.6: Determining Concavity of Functions over Their Domains — part (b)
• TOPIC 4.7: Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms — part (c)
• TOPIC 5.5: Using the Candidates Test to Determine Absolute (Global) Extrema — part (d)
• TOPIC 5.6: Determining Concavity of Functions over Their Domains — part (b)
• TOPIC 4.7: Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms — part (c)
• TOPIC 5.5: Using the Candidates Test to Determine Absolute (Global) Extrema — part (d)
▶️ Answer/Explanation
(a) Relative min/max at x = 6
f′(x) > 0 on (2, 6) and f′(x) > 0 on (6, 8).The sign of f′ does not change at x = 6 → the function is increasing through x = 6.
Conclusion: Neither a relative minimum nor a relative maximum at x = 6.
(b) Concavity
f is concave down where f″(x) < 0 ⇔ where f′ is decreasing.From the graph, f′ decreases on (-2, 0) and (4, 6).
Intervals: (-2, 0) and (4, 6).
(c) Limit \(\displaystyle \lim_{x\to 2}\frac{6f(x)-3x}{x^2-5x+6}\)
Continuity at x = 2: \(\lim_{x\to2} (6f(x)-3x) = 6f(2)-3\cdot2 = 0\), \(\lim_{x\to2}(x^2-5x+6)=0\).Indeterminate form 0/0 → use L’Hôpital’s Rule:
\[ \lim_{x\to2}\frac{6f'(x)-3}{2x-5} = \frac{6f'(2)-3}{-1}. \] From the graph, \(f'(2)=0\). Hence the limit is \(\displaystyle \boxed{3}\).
(d) Absolute minimum of \(f\) on \([-2,8]\)
Candidates: endpoints x = -2, 8 and critical points where f′ = 0 → x = -1, 2, 6.Use \(f(2)=1\) and \(f(x) = f(2) + \int_{2}^{x} f'(t)\,dt\). Compute signed areas from the graph:
• \(\displaystyle \int_{-2}^{-1} f’ = +\tfrac12(1)(2) = +1\) (triangle above axis).
• \(\displaystyle \int_{-1}^{0} f’ = -\tfrac12(1)(2) = -1\) (triangle below axis).
• \(\displaystyle \int_{0}^{2} f’ = -\tfrac12(2)(2) = -2\).
⇒ \(\displaystyle \int_{-2}^{2} f’ = -2\). Hence \(f(-2)=f(2)-(-2)=\boxed{3}\) and \(f(-1)=f(-2)+1=\boxed{4}\).
• \(\displaystyle \int_{2}^{4} f’ = +\tfrac12(2)(2) = +2\).
• \(\displaystyle \int_{4}^{6} f’\) (upper semicircle segment): area = rectangle \(2\times2\) minus quarter–circle of radius 2 → \(4-\pi\).
⇒ \(f(6)=f(2)+2+(4-\pi)=\boxed{7-\pi}\approx 3.859.\)
• By symmetry, \(\displaystyle \int_{6}^{8} f’ = 4-\pi\).
⇒ \(f(8)=f(6)+(4-\pi)=\boxed{11-2\pi}\approx 4.716.\)
Compare candidate values: \(f(-2)=3,\; f(-1)=4,\; f(2)=1,\; f(6)=7-\pi\approx 3.859,\; f(8)=11-2\pi\approx 4.716.\)• \(\displaystyle \int_{-1}^{0} f’ = -\tfrac12(1)(2) = -1\) (triangle below axis).
• \(\displaystyle \int_{0}^{2} f’ = -\tfrac12(2)(2) = -2\).
⇒ \(\displaystyle \int_{-2}^{2} f’ = -2\). Hence \(f(-2)=f(2)-(-2)=\boxed{3}\) and \(f(-1)=f(-2)+1=\boxed{4}\).
• \(\displaystyle \int_{2}^{4} f’ = +\tfrac12(2)(2) = +2\).
• \(\displaystyle \int_{4}^{6} f’\) (upper semicircle segment): area = rectangle \(2\times2\) minus quarter–circle of radius 2 → \(4-\pi\).
⇒ \(f(6)=f(2)+2+(4-\pi)=\boxed{7-\pi}\approx 3.859.\)
• By symmetry, \(\displaystyle \int_{6}^{8} f’ = 4-\pi\).
⇒ \(f(8)=f(6)+(4-\pi)=\boxed{11-2\pi}\approx 4.716.\)
Absolute minimum: \(\boxed{f(2)=1}\) at \(x=2\).