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

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

Guest Sep 18, 2014

#1
+92254
+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.

Melody  Sep 19, 2014
Sort:

#1
+92254
+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.

Melody  Sep 19, 2014

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