Use bounded integral to find local maximum. If the function $f(x)=\int_0^x-\sin t^2 \ dt$ on $-1\leq x\leq3$, then $f$ has local maximum at which $x?$

Question

How do you use a bounded integral to find a local maximum? I didn't know this was even possible.

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The particular multiple choice problem I'm working with is

If the function f is defined by $f(x)=\int_0^x -\sin t^2 \ dt$ on the closed interval $ -1 \leq x \leq 3$, then $f$ has a local maximum at $x = ?$

a. $-1.084$
b. $0$
c. $1.772$
d. $2.171$
e. $2.507$

Answer 1

To answer your question without answering your question directly, you would use the Fundamental Theorem of Calculus. Since $-\sin(x^2)$ is continuous for all values of $x$, then the FTC says that $$f'(x)=-\sin(x^2)$$ Which means that if $f'(c)=0$, then $f$ has an extremum at $c$. So what values of $c$ in $[-1,3]$ will make this true?