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)})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

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}})

p (a, c) = a + 2 + c

f (x) = 40 e^{2 x} - 5 e^{x}

f (x) = x + \frac{49}{x}, 2 \leq x \leq 12

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

f (x, y) = x^{2} + x y - y^{2} - y

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)})