What is the extreme f(x)=x^4-12x^2+36 ?

The extreme f(x)=x^4-12x^2+36 is Minimum(-sqrt(6),0),Maximum(0,36),Minimum(sqrt(6),0)

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extreme f(x)=x^4-12x^2+36

extreme points

f (x) = x^{4} - 12 x^{2} + 36

Minimum (- 6, 0), Maximum (0, 36), Minimum (6, 0)

Solution steps

f^{'} (x) = 4 x^{3} - 24 x

Find intervals:

Decreasing

: - \infty < x < - 6,

Increasing

: - 6 < x < 0,

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = - 6

into

x^{4} - 12 x^{2} + 36 : 0

Minimum (- 6, 0)

Plug

x = 0

into

x^{4} - 12 x^{2} + 36 : 36

Maximum (0, 36)

Plug

x = 6

into

x^{4} - 12 x^{2} + 36 : 0

Minimum (6, 0)

Minimum (- 6, 0), Maximum (0, 36), Minimum (6, 0)

f (x) = 1 + cos (x)

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

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

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

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

What is the extreme f(x)=x^4-12x^2+36 ?
The extreme f(x)=x^4-12x^2+36 is Minimum(-sqrt(6),0),Maximum(0,36),Minimum(sqrt(6),0)