What is the extreme f(x)=-(x+2)^5(x-4)^4 ?

The extreme f(x)=-(x+2)^5(x-4)^4 is Saddle(-2,0),Minimum(4/3 ,-409600000/19683),Maximum(4,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=-(x+2)^5(x-4)^4

extreme points

f (x) = - (x + 2)^{5} (x - 4)^{4}

Saddle (- 2, 0), Minimum (\frac{4}{3}, - \frac{409600000}{19683}), Maximum (4, 0)

Solution steps

f^{'} (x) = - 5 (x + 2)^{4} (x - 4)^{4} - 4 (x - 4)^{3} (x + 2)^{5}

Find intervals:

Decreasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < \frac{4}{3},

Increasing

: \frac{4}{3} < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = - 2

into

- (x + 2)^{5} (x - 4)^{4} : 0

Saddle (- 2, 0)

Plug

x = \frac{4}{3}

into

- (x + 2)^{5} (x - 4)^{4} : - \frac{409600000}{19683}

Minimum (\frac{4}{3}, - \frac{409600000}{19683})

Plug

x = 4

into

- (x + 2)^{5} (x - 4)^{4} : 0

Maximum (4, 0)

Saddle (- 2, 0), Minimum (\frac{4}{3}, - \frac{409600000}{19683}), Maximum (4, 0)

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

f (x) = 3 x^{3} + 2 x^{2} - 10 x + 600

3 x^{2} + 3

f (x) = \frac{8}{x ^{2}} + \frac{1}{( 40 - x ) ^{2}}

f (x, y) = ln (4 x + y) - x^{(2)} - 3 y

What is the extreme f(x)=-(x+2)^5(x-4)^4 ?
The extreme f(x)=-(x+2)^5(x-4)^4 is Saddle(-2,0),Minimum(4/3 ,-409600000/19683),Maximum(4,0)