What is the extreme f(x)=3(x^2+4)(x^2+8)^2 ?

The extreme f(x)=3(x^2+4)(x^2+8)^2 is Minimum(0,768)

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extreme f(x)=3(x^2+4)(x^2+8)^2

extreme points

f (x) = 3 (x^{2} + 4) (x^{2} + 8)^{2}

Minimum (0, 768)

Solution steps

f^{'} (x) = 3 (6 x^{5} + 80 x^{3} + 256 x)

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

3 (x^{2} + 4) (x^{2} + 8)^{2} : 768

Minimum (0, 768)

f (x) = \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 2 x + 1

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

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

f (x) = x^{4} - 5 x^{3} + 9 x^{2}

f (x, y) = x y - 2 x - y + 6

What is the extreme f(x)=3(x^2+4)(x^2+8)^2 ?
The extreme f(x)=3(x^2+4)(x^2+8)^2 is Minimum(0,768)