What is the extreme f(x)=x^4-8x^2+a ?

The extreme f(x)=x^4-8x^2+a is Minimum(-2,a-16),Maximum(0,a),Minimum(2,a-16)

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extreme f(x)=x^4-8x^2+a

extreme points

f (x) = x^{4} - 8 x^{2} + a

Minimum (- 2, a - 16), Maximum (0, a), Minimum (2, a - 16)

Solution steps

f^{'} (x) = 4 x^{3} - 16 x

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

x^{4} - 8 x^{2} + a : a - 16

Minimum (- 2, a - 16)

Plug

x = 0

into

x^{4} - 8 x^{2} + a : a

Maximum (0, a)

Plug

x = 2

into

x^{4} - 8 x^{2} + a : a - 16

Minimum (2, a - 16)

Minimum (- 2, a - 16), Maximum (0, a), Minimum (2, a - 16)

7 \cdot 8^{x + 8} + 6

f (x, y) = x^{3} + y^{2} - 12 x - 2 y

f (x) = x^{2} + 81

f (x, y) = \frac{414}{23} x^{\frac{2}{3}} y^{\frac{1}{3}}

f (x) = 5 x - 4 ln (4 x)

What is the extreme f(x)=x^4-8x^2+a ?
The extreme f(x)=x^4-8x^2+a is Minimum(-2,a-16),Maximum(0,a),Minimum(2,a-16)