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'(x) = g(x)
\(f(x)=\int_{1}^{x}g(t)dt+3\)
\(f(-5)=-\int_{-5}^{1}g(t)dt+3\)
\(f(-5)=-\left ( \frac{1}{2} (1)(2)-\frac{1}{2} (1)(3)-(3)(3)\right )+3\)
\(f(-5)=-\left ( 1-\frac{3}{2}-9\right )+3\)
\(f(-5)=8+\frac{3}{2}+3\)
\(f(-5)=11+\frac{3}{2}\)
(b)
\(\int_{1}^{6}g(x)dx=\int_{1}^{3}g(x)dx+\int_{3}^{6}g(x)dx\)
\(\int_{1}^{6}g(x)dx=(2)(2)+\int_{3}^{6}2(x-4)^{2}dx\)
\(\int_{1}^{6}g(x)dx=4+\left ( \frac{2}{3}(x-4)^{3} \right )|_{3}^{6}\)
\(\int_{1}^{6}g(x)dx=4+\left ( \frac{2}{3}(8)- \frac{2}{3}(-1)\right )\)
\(\int_{1}^{6}g(x)dx=4+\frac{16}{3}+\frac{2}{3}\)
\(\int_{1}^{6}g(x)dx=10\)
(c)
On the interval (0, 1) u(4, 6) f is both increasing and concave up since f'(x) = g(x) and g is positive on the that interval meaning f is increasing on that interval, and g is increasing on that interval, meaning f”(x)>0 on that interval, therefore f is concave up on that interval
(d)
f has a point of inflection at x = 4 since f'(x) = g(x) and since a switches from decreasing to increasing at x = 4, therefore f”(x) = 0 at that point and would change signs from (-) to (+) at x = 4, therefore x = 4 is an inflection point.