What is the critical y=x^2e^x ?

The critical y=x^2e^x is x=-2,x=0

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critical y=x^2e^x

critical points

y = x^{2} e^{x}

x = - 2, x = 0

Solution steps

x^{2} e^{x}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 2, x = 0

Identify critical points not in

f (x)

domain

Domain of

x^{2} e^{x} : - \infty < x < \infty

All critical points are in domain

x = - 2, x = 0

f (x) = \frac{x ^{2} + 25}{x ^{2} - 4}

f (x) = x^{2} + 5 x - 9

f (x) = e^{j t} + e^{0.5 j t}

36 - t^{2}

f (x) = \frac{( x ^{2} - 5 x )}{( 1 - x ^{2} )}