Question:
The continuous function f is defined on the interval -4 ≤ x ≤ 3. The graph of f consists of two quarter circles and one line segment, as shown in the figure above. Let \(g(x)=2x+\int_0^x f(t)dt\).
(a) Find g(-3) . Find g'(x) and evaluate g'(-3) .
(b) Determine the x-coordinate of the point at which g has an absolute maximum on the interval -4 ≤ x ≤ 3. Justify your answer.
(c) Find all values of x on the interval -4 ≤ x ≤ 3 for which the graph of g has a point of inflection. Give a reason for your answer.
(d) Find the average rate of change of f on the interval-4 ≤ x ≤ 3. There is no point c, -4 < c < 3, for which f'(c) is equal to that average rate of change. Explain why this statement does not contradict the Mean Value Theorem.
▶️Answer/Explanation
Ans:
(a)
\(g(-3)=2(-3)+\int_{0}^{-3}f(t)dt = -6 -\frac{9\pi }{4}\)
\(g'(x)=\frac{d}{dx}\left ( 2x + \int_{0}^{x}f(t)dt \right )=2+f(x)\)
g'(-3) = 2 + f(-3) = 2 + 0 = 2
(b)
g'(x) = 0 2 + f(x) = 0
g'(x) f(x) = -2
x = 5/2
\(g(-4)=-8+\int_{0}^{-4}f(t)dt\)
= -8 – 2π
\(g(5/2)= 5+\int^{\frac{5}{2}}_{0} f(t)dt = + \frac{5}{4}\)
\(g(3)= 6+\int_{0}^{3}f(t)dt = 6\)
x = 5/2 , because g’ going from t to – proves it as the only relative maximum and g(5/2) is greater than g at either endpoint.
(c)
f'(x) g”(x) = d/dx (g(x)) = f'(x)
The only point of inflection for g is at x = 0, since f'(x), which is equivalent to g”, only changes signs at x = 0 on the interval -4 ≤ x ≤ 3
(d)
Avg. Rate of change = \(\frac{f(3)-f(-4)}{3-4}\)
\(\frac{-3-1}{3+4}=\frac{-2}{7}\)
Because mean value Theorem only applies when the function is continuous And differentiable on the interval, which doesn’t apply here since f(x) isn’t differentiable at x = 0.