What is the extreme f(x)=x^4-32x^2+6 ?

The extreme f(x)=x^4-32x^2+6 is Minimum(-4,-250),Maximum(0,6),Minimum(4,-250)

Popular >

extreme f(x)=x^4-32x^2+6

extreme points

f (x) = x^{4} - 32 x^{2} + 6

Minimum (- 4, - 250), Maximum (0, 6), Minimum (4, - 250)

Solution steps

f^{'} (x) = 4 x^{3} - 64 x

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 4

into

x^{4} - 32 x^{2} + 6 : - 250

Minimum (- 4, - 250)

Plug

x = 0

into

x^{4} - 32 x^{2} + 6 : 6

Maximum (0, 6)

Plug

x = 4

into

x^{4} - 32 x^{2} + 6 : - 250

Minimum (4, - 250)

Minimum (- 4, - 250), Maximum (0, 6), Minimum (4, - 250)

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

f (x, y) = 2^{2} x^{4} + 2^{2} y^{4} - 4 x y + 16

f (x) = I x - 3 I - 2

f (x) = - x^{4} - 4 x^{3} + 7

f (x) = x^{3} + 3 x

What is the extreme f(x)=x^4-32x^2+6 ?
The extreme f(x)=x^4-32x^2+6 is Minimum(-4,-250),Maximum(0,6),Minimum(4,-250)