What is the extreme y=x^4-8x^2+5 ?

The extreme y=x^4-8x^2+5 is Minimum(-2,-11),Maximum(0,5),Minimum(2,-11)

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extreme y=x^4-8x^2+5

extreme points

y = x^{4} - 8 x^{2} + 5

Minimum (- 2, - 11), Maximum (0, 5), Minimum (2, - 11)

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} + 5 : - 11

Minimum (- 2, - 11)

Plug

x = 0

into

x^{4} - 8 x^{2} + 5 : 5

Maximum (0, 5)

Plug

x = 2

into

x^{4} - 8 x^{2} + 5 : - 11

Minimum (2, - 11)

Minimum (- 2, - 11), Maximum (0, 5), Minimum (2, - 11)

f (x) = (x - 7) e^{x}

y (x, t) = 3 x + 2 a t (3.4)

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

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

y = x^{3} - 4 x^{2}

What is the extreme y=x^4-8x^2+5 ?
The extreme y=x^4-8x^2+5 is Minimum(-2,-11),Maximum(0,5),Minimum(2,-11)