What is the extreme f(x)=x^4-62x^2+120x+9 ?

The extreme f(x)=x^4-62x^2+120x+9 is Minimum(-6,-1647),Maximum(1,68),Minimum(5,-316)

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extreme f(x)=x^4-62x^2+120x+9

extreme points

f (x) = x^{4} - 62 x^{2} + 120 x + 9

Minimum (- 6, - 1647), Maximum (1, 68), Minimum (5, - 316)

Solution steps

f^{'} (x) = 4 x^{3} - 124 x + 120

Find intervals:

Decreasing

: - \infty < x < - 6,

Increasing

: - 6 < x < 1,

Decreasing

: 1 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = - 6

into

x^{4} - 62 x^{2} + 120 x + 9 : - 1647

Minimum (- 6, - 1647)

Plug

x = 1

into

x^{4} - 62 x^{2} + 120 x + 9 : 68

Maximum (1, 68)

Plug

x = 5

into

x^{4} - 62 x^{2} + 120 x + 9 : - 316

Minimum (5, - 316)

Minimum (- 6, - 1647), Maximum (1, 68), Minimum (5, - 316)

f (x) = 9 x^{2} - 36 x + 27

4 x^{2}

f (x, y) = - 5 y^{2} - 6 x^{3} + 12 x^{2}

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

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

What is the extreme f(x)=x^4-62x^2+120x+9 ?
The extreme f(x)=x^4-62x^2+120x+9 is Minimum(-6,-1647),Maximum(1,68),Minimum(5,-316)