What is the extreme f(x)=x^4-32x^2+10 ?

The extreme f(x)=x^4-32x^2+10 is Minimum(-4,-246),Maximum(0,10),Minimum(4,-246)

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extreme f(x)=x^4-32x^2+10

extreme points

f (x) = x^{4} - 32 x^{2} + 10

Minimum (- 4, - 246), Maximum (0, 10), Minimum (4, - 246)

Solution steps

f^{'} (x) = 4 x^{3} - 64 x

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 4

into

x^{4} - 32 x^{2} + 10 : - 246

Minimum (- 4, - 246)

Plug

x = 0

into

x^{4} - 32 x^{2} + 10 : 10

Maximum (0, 10)

Plug

x = 4

into

x^{4} - 32 x^{2} + 10 : - 246

Minimum (4, - 246)

Minimum (- 4, - 246), Maximum (0, 10), Minimum (4, - 246)

f (x) = x^{3} + y^{3} + 3 x^{2} - 9 y^{2} - 1

\frac{x ^{2}}{x ^{2} + 243}

f (x) = 6 sin^{2} (x) + 6 sin (x)

f (x) = 3^{2} - 6 x - 9 = 0

36 x^{5} + 540 x^{4} + 2100 x^{3} - 30

What is the extreme f(x)=x^4-32x^2+10 ?
The extreme f(x)=x^4-32x^2+10 is Minimum(-4,-246),Maximum(0,10),Minimum(4,-246)