Question
(a) Topic-5.4 Using the First Derivative Test to Determine Relative (Local) Extrema
(b) Topic-8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts
(c) Topic-5.6 Determining Concavity of Functions over Their Domains
(d) Topic-2.5 Applying the Power Rule
The figure above shows the graph of f’, the derivative of a twice-differentiable function f, on the closed interval 0 ≤ x ≤ 8. The graph of f’ has horizontal tangent lines at x = 1, x = 3, and x = 5. The areas of the regions between the graph of f’ and the x-axis are labeled in the figure. The function f is defined for all real numbers and satisfies f (8) = 4.
(a) Find all values of x on the open interval 0 < x < 8 for which the function f has a local minimum. Justify your answer.
(b) Determine the absolute minimum value of f on the closed interval 0 ≤ x ≤ 8. Justify your answer.
(c) On what open intervals contained in 0 < x < 8 is the graph of f both concave down and increasing? Explain your reasoning.
(d) The function g is defined by \(g(x)=(f(x))^{3}\), if \(f(3)= -\frac{5}{2},\) find the slope of the line tangent to the graph of g at x = 3.
▶️Answer/Explanation
\(
\textbf{4(a) } x = 6 \text{ is the only critical point at which } f’ \text{ changes sign from negative to positive. Therefore, } f \text{ has a local minimum at } x = 6.
\)
\(
\textbf{4(b)} \text{From part (a), the absolute minimum occurs either at } x = 6 \text{ or at an endpoint.}
\)
\(
f(0) = f(8) + \int_{8}^{0} f'(x) \, dx
\)
\(
= f(8) – \int_{0}^{8} f'(x) \, dx = 4 – 12 = -8
\)
\(
f(6) = f(8) + \int_{8}^{6} f'(x) \, dx
\)
\(
= f(8) – \int_{6}^{8} f'(x) \, dx = 4 – 7 = -3
\)
\(
f(8) = 4
\)
\(
\text{The absolute minimum value of } f \text{ on the closed interval } [0, 8] \text{ is } -8.
\)
\(
\textbf{4(c)} \text{The graph of } f \text{ is concave down and increasing on } 0 < x < 1 \text{ and } 3 < x < 4, \text{ because } f’ \text{ is decreasing and positive on these intervals.}
\)
\(
\textbf{4(d) } g'(x) = 3[f(x)]^{2} \cdot f'(x)
\)
\(
g'(3) = 3[f(3)]^{2} \cdot f'(3) = 3 \left( -\frac{5}{2} \right)^{2} \cdot 4 = 75
\)