What is the extreme f(x)=x^4-32x^2+9 ?

The extreme f(x)=x^4-32x^2+9 is Minimum(-4,-247),Maximum(0,9),Minimum(4,-247)

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extreme f(x)=x^4-32x^2+9

extreme points

f (x) = x^{4} - 32 x^{2} + 9

Minimum (- 4, - 247), Maximum (0, 9), Minimum (4, - 247)

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} + 9 : - 247

Minimum (- 4, - 247)

Plug

x = 0

into

x^{4} - 32 x^{2} + 9 : 9

Maximum (0, 9)

Plug

x = 4

into

x^{4} - 32 x^{2} + 9 : - 247

Minimum (4, - 247)

Minimum (- 4, - 247), Maximum (0, 9), Minimum (4, - 247)

f (x) = sin (\frac{2 π}{3}) x

f (x) = \frac{5 x + 7}{x}

3 cos (2 x + π)

y = x + 8

What is the extreme f(x)=x^4-32x^2+9 ?
The extreme f(x)=x^4-32x^2+9 is Minimum(-4,-247),Maximum(0,9),Minimum(4,-247)