Find the absolute extrema of f(x) = e^{x^2+2x}f (

Ответ:

jekson1252

11.01.2022

$f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}$

has critical points where the derivative is 0:

$2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1$

The second derivative is

$f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}$

and $f''(-1)=\frac2e0$ , which indicates a local minimum at x=-1 with a value of $f(-1)=\frac1e$ .

At the endpoints of [-2, 2], we have f(-2)=1 and f(2)=e^8 , so that has an absolute minimum of $\frac1e$ and an absolute maximum of e^8 on [-2, 2].

So we have

$\dfrac1e\le f(x)\le e^8$

$\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx$

$\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}$

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