Question
The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for −5 ≤ x < 3, and g (x) = 2(x −4 ) 2 for 3 ≤ x ≤ 6.
(a) If f (1) = 3, what is the value of f(−5) ?
(b) Evaluate \(\int_{1}^{6}g(x)dx.\)
(c) For −5 < x < 6, on what open intervals, if any, is the graph of f both increasing and concave up? Give a reason for your answer.
(d) Find the x-coordinate of each point of inflection of the graph of f. Give a reason for your answer.
Answer/Explanation
Ans:
(a) \(f(-5)=f(1)+\int_{1}^{-5}g(x)dx=f(1)-\int_{-5}^{1}g(x)dx\)
\(=3-\left ( -9-\frac{3}{2}+1 \right )=3-\left ( -\frac{19}{2} \right )=\frac{25}{2}\)
(b) \(\int_{1}^{6}g(x)dx=\int_{1}^{3}g(x)dx+\int_{3}^{6}g(x)dx\)
\(=\int_{1}^{3}2dx+\int_{3}^{6}2(x-4)^{2}dx\)
\(=4+_{x=3}\left [ \frac{2}{3}(x-4)^{3} \right ]^{x=6}=4+\frac{16}{3}-\left ( -\frac{2}{3} \right )=10\)
(c) The graph of f is increasing and concave up on 0 < x < 1 and 4 < x < 6 because f ‘(x) =g(x)> 0 and f ‘(x) = g(x) is increasing on those intervals.
(d) The graph of f has a point of inflection at x = 4 because f'(x) = g(x) changes from decreasing to increasing at x = 4.