What is the extreme f(x)=x^4-98x^2+1 ?

The extreme f(x)=x^4-98x^2+1 is Minimum(-7,-2400),Maximum(0,1),Minimum(7,-2400)

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extreme f(x)=x^4-98x^2+1

extreme points

f (x) = x^{4} - 98 x^{2} + 1

Minimum (- 7, - 2400), Maximum (0, 1), Minimum (7, - 2400)

Solution steps

f^{'} (x) = 4 x^{3} - 196 x

Find intervals:

Decreasing

: - \infty < x < - 7,

Increasing

: - 7 < x < 0,

Decreasing

: 0 < x < 7,

Increasing

: 7 < x < \infty

Plug

x = - 7

into

x^{4} - 98 x^{2} + 1 : - 2400

Minimum (- 7, - 2400)

Plug

x = 0

into

x^{4} - 98 x^{2} + 1 : 1

Maximum (0, 1)

Plug

x = 7

into

x^{4} - 98 x^{2} + 1 : - 2400

Minimum (7, - 2400)

Minimum (- 7, - 2400), Maximum (0, 1), Minimum (7, - 2400)

f (x) = x ∣ x + 4 ∣

f (x) = - 3 (x - 1)^{2} (x^{2} - 4)

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

f (x) = \frac{x ^{2} - 9}{x ^{2} - 4 x - 21}

x^{3} + 3 x^{2} + 5 x + 7

What is the extreme f(x)=x^4-98x^2+1 ?
The extreme f(x)=x^4-98x^2+1 is Minimum(-7,-2400),Maximum(0,1),Minimum(7,-2400)