What is the critical f(x)=e^x(2x^2+x-8) ?

The critical f(x)=e^x(2x^2+x-8) is x=-7/2 ,x=1

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critical f(x)=e^x(2x^2+x-8)

critical points

f (x) = e^{x} (2 x^{2} + x - 8)

x = - \frac{7}{2}, x = 1

Solution steps

e^{x} (2 x^{2} + x - 8)

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{7}{2}, x = 1

Identify critical points not in

f (x)

domain

Domain of

e^{x} (2 x^{2} + x - 8) : - \infty < x < \infty

All critical points are in domain

x = - \frac{7}{2}, x = 1

\frac{x ^{3}}{3} + \frac{3}{2} y^{2} + 3 x y + 2 x + 6

f (x) = 3 x^{3} - 3 x^{2} - 3 x - 1

f (x) = 2 x^{3} + 3 x^{2} - 36 x + 5

f (x) = 2 x^{3} + 3 x^{2} - 36 x + 4

What is the critical f(x)=e^x(2x^2+x-8) ?
The critical f(x)=e^x(2x^2+x-8) is x=-7/2 ,x=1