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

The extreme f(x)=x^4e^x-2 is Maximum(-4,(256)/(e^4)-2),Minimum(0,-2)

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

extreme points

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

Maximum (- 4, \frac{256}{e ^{4}} - 2), Minimum (0, - 2)

Solution steps

f^{'} (x) = e^{x} x^{4} + 4 e^{x} x^{3}

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 4

into

x^{4} e^{x} - 2 : \frac{256}{e ^{4}} - 2

Maximum (- 4, \frac{256}{e ^{4}} - 2)

Plug

x = 0

into

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

Minimum (0, - 2)

Maximum (- 4, \frac{256}{e ^{4}} - 2), Minimum (0, - 2)

f (x) = x^{4} - 8 x^{3}

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

f (x) = - x^{2} - 9 x - 20

y = lo g_{10} (x)

f (x) = ln (9 - x^{2})

What is the extreme f(x)=x^4e^x-2 ?
The extreme f(x)=x^4e^x-2 is Maximum(-4,(256)/(e^4)-2),Minimum(0,-2)