What is the extreme f(x)=x^4-18x^2+1 ?

The extreme f(x)=x^4-18x^2+1 is Minimum(-3,-80),Maximum(0,1),Minimum(3,-80)

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extreme f(x)=x^4-18x^2+1

extreme points

f (x) = x^{4} - 18 x^{2} + 1

Minimum (- 3, - 80), Maximum (0, 1), Minimum (3, - 80)

Solution steps

f^{'} (x) = 4 x^{3} - 36 x

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

x^{4} - 18 x^{2} + 1 : - 80

Minimum (- 3, - 80)

Plug

x = 0

into

x^{4} - 18 x^{2} + 1 : 1

Maximum (0, 1)

Plug

x = 3

into

x^{4} - 18 x^{2} + 1 : - 80

Minimum (3, - 80)

Minimum (- 3, - 80), Maximum (0, 1), Minimum (3, - 80)

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

f (x, y) = (9 + x y) (x + y)

f (x) = 3 x^{5} - 5 x^{3} + 1

f (x) = x^{3} + 3 x^{2} - 9 x + 5

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

What is the extreme f(x)=x^4-18x^2+1 ?
The extreme f(x)=x^4-18x^2+1 is Minimum(-3,-80),Maximum(0,1),Minimum(3,-80)