What is the extreme f(x)=x^4+16x^3-6 ?

The extreme f(x)=x^4+16x^3-6 is Minimum(-12,-6918),Saddle(0,-6)

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extreme f(x)=x^4+16x^3-6

extreme points

f (x) = x^{4} + 16 x^{3} - 6

Minimum (- 12, - 6918), Saddle (0, - 6)

Solution steps

f^{'} (x) = 4 x^{3} + 48 x^{2}

Find intervals:

Decreasing

: - \infty < x < - 12,

Increasing

: - 12 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 12

into

x^{4} + 16 x^{3} - 6 : - 6918

Minimum (- 12, - 6918)

Plug

x = 0

into

x^{4} + 16 x^{3} - 6 : - 6

Saddle (0, - 6)

Minimum (- 12, - 6918), Saddle (0, - 6)

f (x) = \frac{x ^{2} + 1}{3 x + 0}

\frac{2 x + 1}{x - 1}

f (x) = 7 x - x^{3}

x^{7} ln (x)

What is the extreme f(x)=x^4+16x^3-6 ?
The extreme f(x)=x^4+16x^3-6 is Minimum(-12,-6918),Saddle(0,-6)