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Determine the local extrema of $F(x)=\int^{x^2}_1\frac{\sin t}{2+e^t}\,dt$.
Prove that $|F(x)|\leq|x-1|$ for all $x$.

$$F'(x)=\frac{2x\sin x}{2+e^x}\mbox{,}$$ so that the set of all points at which $F$ may attains its local extremum are $n\pi$, $n\in\mathbb{Z}$.

I don't know how to proceed any further.

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  • $\begingroup$ Have you, firstly, worked out the derivative of this function with respect to $x$? $\endgroup$
    – Prometheus
    Commented Nov 22, 2021 at 8:42
  • $\begingroup$ You should have some $x^2$ terms in your derivative. $\endgroup$
    – user317176
    Commented Nov 22, 2021 at 8:49
  • $\begingroup$ $F(1) = 0$ and $F'(x) < 1$ for all $x.$ So $F(x) \le \int_1^x t \ dt$ $\endgroup$
    – user317176
    Commented Nov 22, 2021 at 8:52
  • $\begingroup$ I see the second problem follows from the mean value theorem now. $\endgroup$
    – user912011
    Commented Nov 22, 2021 at 8:56

1 Answer 1

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The maybe more than one way to answer your question, however using the mean value theorem would be the most direct way to do it.

We have, $$ |F'(x)|=\left|\dfrac{2x\sin(x^2)}{2+\exp(x^2)}\right|\leq\left|\dfrac{2x}{2+\exp(x^2)}\right|\leq\left| \dfrac{x}{1 + \frac{1}{2}\exp(x^2)}\right|\leq 1$$ The last inequality is valid for all $x\in\mathbb{R}$, because then we have $2x\leq\exp(x^2)$

Given that $F(x)$ is continuous on $[1,x]$ and derivable on $]1,x[$, we can conclude using the MVT: $$ \left|\dfrac{F(x) - F(1)}{x-1} \right|\leq 1$$ Finall, given that $F(1)=0$ we have $$ \left| F(x)\right| \leq \left| x-1 \right|, \forall x\in\mathbb{R}$$


About the local extrema: To find the local extrema of the function $F$, we need to solve the equation $F'(x)=0 $.

$$ F'(x)=0 \Longleftrightarrow \frac{2x\sin x^2}{2+\exp(x^2)}=0$$ Thus, $$ x= 0,\quad \mbox{or }\ \sin(x^2)=0 \Leftrightarrow x^2=n\pi,\ n\in\mathbb{Z}$$

We can than conclude that the local extrema of the function are of the form $\pm \sqrt{n\pi}, n \in \mathbb{N}$

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    $\begingroup$ How to determine the points at which $F$ attains it local maxima and minima? $\endgroup$
    – user912011
    Commented Nov 22, 2021 at 9:27
  • $\begingroup$ $3x^2=0$ when $x=0$ but $0$ is not an extrema of $x^3$ $\endgroup$
    – user912011
    Commented Nov 23, 2021 at 0:58
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    $\begingroup$ Well, that's correct. However, that's not the case here, because the function $2x\sin(x^2)$ changes sign every time it crosses the $y=0$ line. $\endgroup$
    – Varazda
    Commented Nov 23, 2021 at 7:53

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