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Find the max and min of f(x)=x^2e^-x/2, [-1,8]

 Sep 18, 2014

Best Answer 

 #1
avatar+95179 
+5

f(x)=x^2e^-x/2, [-1,8]

 

$$\\f(x)=x^2e^{-\frac{x}{2}}, [-1,8]\\\\
f'(x)=2xe^{-\frac{x}{2}}+\frac{-1}{2}e^{-\frac{x}{2}}x^2\\\\
f'(x)=2xe^{-\frac{x}{2}}-0.5x^2e^{-\frac{x}{2}}\\\\
f'(x)=xe^{-\frac{x}{2}}(2-0.5x)\\\\
$stationary points when $ f'(x)=0\\\\
x=0\;\;or\;\;2=0.5x\\
x=0\;\;or\;\;x=4\\$$

 

So you now need to find the y values for 

 

$$x=0,\;x=4,\;x=-1\;\; and\;\; x=8$$

 

Then you will have your minimum and your maximum values for the given region.

 Sep 19, 2014
 #1
avatar+95179 
+5
Best Answer

f(x)=x^2e^-x/2, [-1,8]

 

$$\\f(x)=x^2e^{-\frac{x}{2}}, [-1,8]\\\\
f'(x)=2xe^{-\frac{x}{2}}+\frac{-1}{2}e^{-\frac{x}{2}}x^2\\\\
f'(x)=2xe^{-\frac{x}{2}}-0.5x^2e^{-\frac{x}{2}}\\\\
f'(x)=xe^{-\frac{x}{2}}(2-0.5x)\\\\
$stationary points when $ f'(x)=0\\\\
x=0\;\;or\;\;2=0.5x\\
x=0\;\;or\;\;x=4\\$$

 

So you now need to find the y values for 

 

$$x=0,\;x=4,\;x=-1\;\; and\;\; x=8$$

 

Then you will have your minimum and your maximum values for the given region.

Melody Sep 19, 2014

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