Question
(a) Topic-5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points
(b) Topic-5.1 Using the Mean Value Theorem
(c) Topic-2.7 Derivatives of \(\cos x\), \(\sin x\), \(e^x\), and \(\ln x\)
(d) Topic-6.7 The Fundamental Theorem of Calculus and Definite Integrals
The twice-differentiable functions f and g are defined for all real numbers x. Values of f, f’, g, and g’ for various values of x are given in the table above.
(a) Find the x-coordinate of each relative minimum of f on the interval [-2, 3] . Justify your answers.
(b) Explain why there must be a value c, for -1 < c < 1, such that f”(c) =0.
(c) The function h is defined by h(x) = In (f(x)). Find h'(3). Show the computations that lead to your answer.
(d) Evaluate \(\int_{-2}^{3}f'(g(x))g'(x)dx.\)
▶️Answer/Explanation
\(
\textbf{5(a) } x = 1 \text{ is the only critical point at which } f’ \text{ changes sign from negative to positive.
Therefore, } f \text{ has a relative minimum at } x = 1.
\)
\(
\textbf{5(b) } f’ \text{ is differentiable } \implies f’ \text{ is continuous on the interval } -1 \leq x \leq 1
\)
\(
\frac{f'(1) – f'(-1)}{1 – (-1)} = \frac{0 – 0}{2} = 0
\)
\(
\text{Therefore, by the Mean Value Theorem, there is at least one value } c, \, -1 < c < 1, \text{ such that } f”(c) = 0.
\)
\(
\textbf{5(c) } h'(x) = \frac{1}{f(x)} \cdot f'(x)
\)
\(
h'(3) = \frac{1}{f(3)} \cdot f'(3) = \frac{1}{7} \cdot \frac{1}{2} = \frac{1}{14}
\)
\(
\textbf{5(d) } \int_{2}^{3} f'(g(x))g'(x) \, dx = \left[ f(g(x)) \right]_{x=2}^{x=3}
\)
\(
= f(g(3)) – f(g(2))
\)
\(
= f(1) – f(-1)
\)
\(
= 2 – 8 = -6
\)