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

Question

3099 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-09T11:50:58

#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#.