What is the extreme f(x)=17x^4-102x^2 ?

The extreme f(x)=17x^4-102x^2 is Minimum(-sqrt(3),-153),Maximum(0,0),Minimum(sqrt(3),-153)

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extreme f(x)=17x^4-102x^2

extreme points

f (x) = 17 x^{4} - 102 x^{2}

Minimum (- 3, - 153), Maximum (0, 0), Minimum (3, - 153)

Solution steps

f^{'} (x) = 68 x^{3} - 204 x

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

17 x^{4} - 102 x^{2} : - 153

Minimum (- 3, - 153)

Plug

x = 0

into

17 x^{4} - 102 x^{2} : 0

Maximum (0, 0)

Plug

x = 3

into

17 x^{4} - 102 x^{2} : - 153

Minimum (3, - 153)

Minimum (- 3, - 153), Maximum (0, 0), Minimum (3, - 153)

f (x) = ln (8 x), x > 0

(1, 5), (2, 3)

(12, 2), (- 5, - 7)

\frac{18}{x ^{2} + 12}

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

What is the extreme f(x)=17x^4-102x^2 ?
The extreme f(x)=17x^4-102x^2 is Minimum(-sqrt(3),-153),Maximum(0,0),Minimum(sqrt(3),-153)