What is the extreme f(x)=e^x(15-x^2) ?

The extreme f(x)=e^x(15-x^2) is Minimum(-5,-(10)/(e^5)),Maximum(3,6e^3)

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extreme f(x)=e^x(15-x^2)

extreme points

f (x) = e^{x} (15 - x^{2})

Minimum (- 5, - \frac{10}{e ^{5}}), Maximum (3, 6 e^{3})

Solution steps

f^{'} (x) = - e^{x} (x - 3) (x + 5)

Find intervals:

Decreasing

: - \infty < x < - 5,

Increasing

: - 5 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = - 5

into

e^{x} (15 - x^{2}) : - \frac{10}{e ^{5}}

Minimum (- 5, - \frac{10}{e ^{5}})

Plug

x = 3

into

e^{x} (15 - x^{2}) : 6 e^{3}

Maximum (3, 6 e^{3})

Minimum (- 5, - \frac{10}{e ^{5}}), Maximum (3, 6 e^{3})

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

f (x) = 10 + 4 x + \frac{9}{x}

f (x) = 8 x^{3} - 17 x^{2} + 12 x + 14

f (x, y) = 3 y^{3} + 7 x y - 9 x^{2} - 5

f (x) = - \frac{cos ( 2 x )}{2} - 2 sin (x)

What is the extreme f(x)=e^x(15-x^2) ?
The extreme f(x)=e^x(15-x^2) is Minimum(-5,-(10)/(e^5)),Maximum(3,6e^3)