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

The critical f(x)=4x^2-e^x is x=-\W_{0}(-1/8),x=-\W_{-1}(-1/8)

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critical f(x)=4x^2-e^x

critical points

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

x = - W_{0} (- \frac{1}{8}), x = - W_{- 1} (- \frac{1}{8})

Solution steps

4 x^{2} - e^{x}

Find where

f^{'} (x)

is equal to zero or undefined

x = - W_{0} (- \frac{1}{8}), x = - W_{- 1} (- \frac{1}{8})

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = - W_{0} (- \frac{1}{8}), x = - W_{- 1} (- \frac{1}{8})

7 - 3 x

(1.7) (9)

(8, 13), (6, - 1)

h (x) = x - 9

f (x) = 5 x^{4} - 10 x^{2} + 8

What is the critical f(x)=4x^2-e^x ?
The critical f(x)=4x^2-e^x is x=-\W_{0}(-1/8),x=-\W_{-1}(-1/8)