What is the extreme f(x)=e^{(x)}(2x^2+x-8) ?

The extreme f(x)=e^{(x)}(2x^2+x-8) is Maximum(-7/2 ,(13)/(e^{7/2)}),Minimum(1,-5e)

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

extreme points

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

Maximum (- \frac{7}{2}, \frac{13}{e ^{\frac{7}{2}}}), Minimum (1, - 5 e)

Solution steps

f^{'} (x) = e^{x} (x - 1) (2 x + 7)

Find intervals:

Increasing

: - \infty < x < - \frac{7}{2},

Decreasing

: - \frac{7}{2} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - \frac{7}{2}

into

e^{x} (2 x^{2} + x - 8) : \frac{13}{e ^{\frac{7}{2}}}

Maximum (- \frac{7}{2}, \frac{13}{e ^{\frac{7}{2}}})

Plug

x = 1

into

e^{x} (2 x^{2} + x - 8) : - 5 e

Minimum (1, - 5 e)

Maximum (- \frac{7}{2}, \frac{13}{e ^{\frac{7}{2}}}), Minimum (1, - 5 e)

What is the extreme f(x)=e^{(x)}(2x^2+x-8) ?
The extreme f(x)=e^{(x)}(2x^2+x-8) is Maximum(-7/2 ,(13)/(e^{7/2)}),Minimum(1,-5e)