What is the extreme f(x)=x^4-18x^2+5 ?

The extreme f(x)=x^4-18x^2+5 is Minimum(-3,-76),Maximum(0,5),Minimum(3,-76)

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extreme f(x)=x^4-18x^2+5

extreme points

f (x) = x^{4} - 18 x^{2} + 5

Minimum (- 3, - 76), Maximum (0, 5), Minimum (3, - 76)

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} + 5 : - 76

Minimum (- 3, - 76)

Plug

x = 0

into

x^{4} - 18 x^{2} + 5 : 5

Maximum (0, 5)

Plug

x = 3

into

x^{4} - 18 x^{2} + 5 : - 76

Minimum (3, - 76)

Minimum (- 3, - 76), Maximum (0, 5), Minimum (3, - 76)

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

f (x) = - 2 (x - 3) (x + 5)

x^{3} - 2 x^{2} - 15 x + 5, - 2 \leq x \leq 0

x^{2} + y x + y^{2}

\frac{7 x - 6}{2 x - 8}

What is the extreme f(x)=x^4-18x^2+5 ?
The extreme f(x)=x^4-18x^2+5 is Minimum(-3,-76),Maximum(0,5),Minimum(3,-76)