What is the extreme f(x)=e^{x^3-6x^2} ?

The extreme f(x)=e^{x^3-6x^2} is Maximum(0,1),Minimum(4, 1/(e^{32)})

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extreme f(x)=e^{x^3-6x^2}

extreme

Maximum (0, 1), Minimum (4, \frac{1}{e ^{32}})

Solution steps

f^{'} (x) = e^{x^{3} - 6 x^{2}} (3 x^{2} - 12 x)

Find intervals:

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

e^{x^{3} - 6 x^{2}} : 1

Maximum (0, 1)

Plug

x = 4

into

e^{x^{3} - 6 x^{2}} : \frac{1}{e ^{32}}

Minimum (4, \frac{1}{e ^{32}})

Maximum (0, 1), Minimum (4, \frac{1}{e ^{32}})

What is the extreme f(x)=e^{x^3-6x^2} ?
The extreme f(x)=e^{x^3-6x^2} is Maximum(0,1),Minimum(4, 1/(e^{32)})