Answer to Question #147653 in Calculus for liam donohue

Question #147653

Let f(x)=e7x2f(x)=e^{-7x^2} .

Then f(x) has a relative minimum at

x=


a relative maximum at

x=


and inflection points at

x=


and at

x=


Write DNE if any of the above do not exist. Write the inflection points (if any) in numerical order, smallest first.


1
Expert's answer
2020-12-10T20:24:24-0500

Solution:Domain of given function is <x<-\infty \:<x<\infty

Combinethecriticalpoint(s):x=0withthedomain\mathrm{Combine\:the\:critical\:point\left(s\right):}\:x=0\:\mathrm{with\:the\:domain}

f(x)=e7x2f(x)=e^{-7x^2}

f(x)=14xe7x2f'(x)=-14xe^{-7x^2}

14xe7x2=0-14xe^{-7x^2}=0

x=0x=0


Thefunctionmonotoneintervalsare:\mathrm{The\:function\:monotone\:intervals\:are:} <x<0-\infty \:<x<0 , 0<x<0<x<\infty \:

We should check the sign f(x)=14xe7x2f'(x)=-14xe^{-7x^2} at each monotone interval

SummaryofthemonotoneintervalsbehaviorSummary\:of\:the\:monotone\:intervals\:behavior :


1) <x<0increasing+-\infty \:<x<0\Rightarrow increasing\Rightarrow+

2) 0<x<0<x<\infty \: decreasing\Rightarrow decreasing\Rightarrow -

3) x=0max0x=0 \Rightarrow max \Rightarrow0

Plugtheextremepointx=0intoe7x2y=1\mathrm{Plug\:the\:extreme\:point}\:x=0\:\mathrm{into}\:e^{-7x^2}\quad \Rightarrow \quad \:y=1


f(x) has a relative minimum at:DNErelative maximum at:

maximum(0,1)\mathrm{maximum}\left(0,\:1\right)


inflection points at:

\mathrm{If\:}f\:''\left(x\right)>0\mathrm{\:then\:}f\left(x\right)\mathrm{\:concave\:upwards.}

\mathrm{If\:}f\:''\left(x\right)<0\mathrm{\:then\:}f\left(x\right)\mathrm{\:concave\:downwards.}


f(x)=0f''(x)=0


x=(1414,1e)\left(-\frac{\sqrt{14}}{14},\:\frac{1}{\sqrt{e}}\right)


and at

x=(1414,1e)\left(\frac{\sqrt{14}}{14},\:\frac{1}{\sqrt{e}}\right)



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