Question
(a)-Topic-5.9 Connecting a Function, Its First Derivative, and Its Second Derivative
(b)-Topic-5.6 Determining Concavity of Functions over Their 2 Domains
(c)-Topic-4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms
(d)-Topic-5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points
4. The function f is defined on the closed interval [−2, 8] and satisfies f(2) = 1. The graph of f ‘ , the derivative of f , consists of two line segments and a semicircle, as shown in the figure.
(a) Does f have a relative minimum, a relative maximum, or neither at x = 6 ? Give a reason for your answer.
(b) On what open intervals, if any, is the graph of f concave down? Give a reason for your answer.
(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 the closed interval [−2, 8]. Justify your answer.
▶️Answer/Explanation
4(a) Does f have a relative minimum, a relative maximum, or neither at x = 6 ? Give a reason for your answer.
f ′(x ) > 0 on (2, 6) and f ′( x) > 0 on (6, 8 ).
f ′(x ) does not change sign at x = 6, so there is neither a relative maximum nor a relative minimum at this location.
4(b) On what open intervals, if any, is the graph of f concave down? Give a reason for your answer.The graph of f is concave down on ( 2, 0) − and (4, 6) because f ′ is decreasing on these intervals.
4(c) Find the value of \(\lim_{ x\to 2}\frac{6f(x)-3x}{x^{2}-5x+6}\),or show that it does not exist. Justify your answer
Because f is differentiable at x = 2, f is continuous at x = 2,
so \(\displaystyle \lim_{x \to 2}f(x)=f(2)=1\)
\(\displaystyle \lim_{x \to 2}(6f(x)-3x)=6.1-3.2=0\)
\(\displaystyle \lim_{x \to 2}(x^{2}-5x+6)=0\)
Because \(\lim_{ x\to 2}\frac{6f(x)-3x}{x^{2}-5x+6}\) is of indeterminate form \(\frac{0}{0}\)
L’Hospital’s Rule can be applied.
Using L’Hospital’s Rule,
\(\displaystyle \lim_{x \to 2}\frac{6f(x)-3x}{x^{2}-5x+6}=\displaystyle \lim_{x \to 2}\frac{6f'(x) – 3}{2.2 – 5} = 3.\)
4(d) Find the absolute minimum value of f on the closed interval [−2, 8 .] Justify your answer.
f′(x ) = 0 ⇒ x =− 1, x = 2, x = 6
The function f is continuous on [−2, 8 ,] so the candidates for the location of an absolute minimum for f are x = −2, x = −1, x = 2, x = 6, and x = 8.
The absolute minimum value of f is f (2)= 1 .