What is the critical f(x)=x^2+e^{16x} ?

The critical f(x)=x^2+e^{16x} is x=-(\W_{0}(128))/(16)

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critical f(x)=x^2+e^{16x}

critical points

f (x) = x^{2} + e^{16 x}

x = - \frac{W _{0} ( 128 )}{16}

Solution steps

x^{2} + e^{16 x}

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{W _{0} ( 128 )}{16}

Identify critical points not in

f (x)

domain

Domain of

x^{2} + e^{16 x} : - \infty < x < \infty

All critical points are in domain

x = - \frac{W _{0} ( 128 )}{16}

f (x) = - x + 4 - 1

3 x - 5 y = 15

- 4 x + 3

x 4 - x^{2}

What is the critical f(x)=x^2+e^{16x} ?
The critical f(x)=x^2+e^{16x} is x=-(\W_{0}(128))/(16)