What is the extreme f(x)=e^x(3-x^2) ?

The extreme f(x)=e^x(3-x^2) is Minimum(-3,-6/(e^3)),Maximum(1,2e)

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extreme f(x)=e^x(3-x^2)

extreme points

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

Minimum (- 3, - \frac{6}{e ^{3}}), Maximum (1, 2 e)

Solution steps

f^{'} (x) = - e^{x} (x - 1) (x + 3)

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 3

into

e^{x} (3 - x^{2}) : - \frac{6}{e ^{3}}

Minimum (- 3, - \frac{6}{e ^{3}})

Plug

x = 1

into

e^{x} (3 - x^{2}) : 2 e

Maximum (1, 2 e)

Minimum (- 3, - \frac{6}{e ^{3}}), Maximum (1, 2 e)

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

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

f (x) = x^{3} - 12 x^{2}

f (x) = 2 x^{3} + 3 x^{2} - 336 x

y = x^{4} + 6 x^{3} + 12 x^{2} + 8 x + 10

What is the extreme f(x)=e^x(3-x^2) ?
The extreme f(x)=e^x(3-x^2) is Minimum(-3,-6/(e^3)),Maximum(1,2e)