What is the extreme f(x)=(x^2+1)(4x^2-1) ?

The extreme f(x)=(x^2+1)(4x^2-1) is Minimum(0,-1)

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extreme f(x)=(x^2+1)(4x^2-1)

extreme points

f (x) = (x^{2} + 1) (4 x^{2} - 1)

Minimum (0, - 1)

Solution steps

f^{'} (x) = 2 x (8 x^{2} + 3)

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

(x^{2} + 1) (4 x^{2} - 1) : - 1

Minimum (0, - 1)

72 y^{2} + x^{2} - x^{2} y

f (x) = x^{4} - 32 x + 3

f (x) = x^{3} - 6 x - 4

f (x) = x^{\frac{2}{3}} (20 - 3 x)

f (x, y) = 3 x^{2} y^{3} - y^{2} - x^{2} + 6

What is the extreme f(x)=(x^2+1)(4x^2-1) ?
The extreme f(x)=(x^2+1)(4x^2-1) is Minimum(0,-1)