What are the absolute extrema of $f (x) = x \frac{e^{x^{2}}}{128} \in [- 5, 16]$ $f(x)= xe^(x^2)/128in [-5,16]$ ?

Question

3580 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-09T11:50:58

$f (16) = \frac{e^{256}}{8} and f (0) = 0$ $f(16)=e^256/8 and f(0)=0$

$f'=(1/128)(xe^(x^2))'=(1/128)((x)'e^(x^2)+(e^(x^2))'x)=(1/128)e^(x^2)(1+2x^2)$

Here, $e^(x^2)>=1 and 1+2x^2>=1$ . So, f'>=1>0#

And so, f is an increasing function in $x in (-oo, oo)$

As $|f(-5)|=(5/128)e^25 < f(16) = e^256/8$ ,

the absolute maximum = $f(16)=e^256/8$ and,

as $|f| >= 0 and f(0) = 0$ ,

the absolute minimum= $f(0)=0$ .

What are the absolute extrema of f(x)= xe^(x^2)/128in [-5,16]f(x)=xex2128∈[−5,16] f(x)= xe^(x^2)/128in [-5,16]?