let g(x)=∫₀ˣ f(t)dt, where the graph of f is shown below. for what value(s) of x does g have a critical…

Explanation:

Step1: Recall the fundamental theorem of calculus

By the fundamental - theorem of calculus, if $g(x)=\int_{0}^{x}f(t)dt$, then $g^\prime(x) = f(x)$.

Step2: Recall the definition of critical points

Critical points of a function $y = g(x)$ occur where $g^\prime(x)=0$ or $g^\prime(x)$ is undefined. Since $g^\prime(x)=f(x)$, we need to find the $x$ - values where $f(x) = 0$.

Step3: Identify $x$ - values from the graph of $f(x)$

Looking at the graph of $f(x)$, we see that $f(x)=0$ at $x=-2,2,8$.

Answer:

-2, 2 and 8