What is the critical f(x)=x^2e^{4x} ?

The critical f(x)=x^2e^{4x} is x=-1/2 ,x=0

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critical f(x)=x^2e^{4x}

critical points

f (x) = x^{2} e^{4 x}

x = - \frac{1}{2}, x = 0

Solution steps

x^{2} e^{4 x}

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{1}{2}, x = 0

Identify critical points not in

f (x)

domain

Domain of

x^{2} e^{4 x} : - \infty < x < \infty

All critical points are in domain

x = - \frac{1}{2}, x = 0

2 x + 6 y = 1, (- 2, 2)

5 x - 2

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

\frac{7}{x + 3}

f (x) = \frac{( 2 - 3 x )}{( 4 x + 6 )}

What is the critical f(x)=x^2e^{4x} ?
The critical f(x)=x^2e^{4x} is x=-1/2 ,x=0