What is the extreme f(x)=e^{x^3-12x} ?

The extreme f(x)=e^{x^3-12x} is Maximum(-2,e^{16}),Minimum(2, 1/(e^{16)})

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extreme f(x)=e^{x^3-12x}

extreme points

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

Maximum (- 2, e^{16}), Minimum (2, \frac{1}{e ^{16}})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

e^{x^{3} - 12 x} : e^{16}

Maximum (- 2, e^{16})

Plug

x = 2

into

e^{x^{3} - 12 x} : \frac{1}{e ^{16}}

Minimum (2, \frac{1}{e ^{16}})

Maximum (- 2, e^{16}), Minimum (2, \frac{1}{e ^{16}})

(x - 8 x^{2})^{\frac{1}{3}}

f (x) = x 4 - x^{2}, - 2 \leq x \leq 2

x^{2} + 8 x + 18

f (x) = x e^{- \frac{x}{2}}

1 - x + 1 + x + 2 x

What is the extreme f(x)=e^{x^3-12x} ?
The extreme f(x)=e^{x^3-12x} is Maximum(-2,e^{16}),Minimum(2, 1/(e^{16)})