What is the extreme f(x)=2x^4-8x^2+6 ?

The extreme f(x)=2x^4-8x^2+6 is Minimum(-sqrt(2),-2),Maximum(0,6),Minimum(sqrt(2),-2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=2x^4-8x^2+6

extreme points

f (x) = 2 x^{4} - 8 x^{2} + 6

Minimum (- 2, - 2), Maximum (0, 6), Minimum (2, - 2)

Solution steps

f^{'} (x) = 8 x^{3} - 16 x

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

2 x^{4} - 8 x^{2} + 6 : - 2

Minimum (- 2, - 2)

Plug

x = 0

into

2 x^{4} - 8 x^{2} + 6 : 6

Maximum (0, 6)

Plug

x = 2

into

2 x^{4} - 8 x^{2} + 6 : - 2

Minimum (2, - 2)

Minimum (- 2, - 2), Maximum (0, 6), Minimum (2, - 2)

f (x) = \frac{x ^{3} + 1}{x ^{2}}

y = \frac{1}{2} x - \frac{3}{2}

\frac{x - 2}{3 x - 4}

y = - 5 x^{2} + 20 x

f (x) = 16 - x^{2} + x + 1

What is the extreme f(x)=2x^4-8x^2+6 ?
The extreme f(x)=2x^4-8x^2+6 is Minimum(-sqrt(2),-2),Maximum(0,6),Minimum(sqrt(2),-2)