What is the extreme f(x)=2x^4-16x^3 ?

The extreme f(x)=2x^4-16x^3 is Saddle(0,0),Minimum(6,-864)

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extreme f(x)=2x^4-16x^3

extreme points

f (x) = 2 x^{4} - 16 x^{3}

Saddle (0, 0), Minimum (6, - 864)

Solution steps

f^{'} (x) = 8 x^{3} - 48 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 0

into

2 x^{4} - 16 x^{3} : 0

Saddle (0, 0)

Plug

x = 6

into

2 x^{4} - 16 x^{3} : - 864

Minimum (6, - 864)

Saddle (0, 0), Minimum (6, - 864)

f (x) = x^{2} - 8 x, 0 \leq x \leq 9

f (x) = 2 - 3 x, 0 \leq x \leq 1

f (x) = (\frac{1}{12} x^{3} + \frac{- 1}{2} x^{2} + 12)

y = x^{2} + 4 x - 7

f (x) = \frac{4}{3} x^{\frac{3}{2}} - \frac{1}{2} x^{2}

What is the extreme f(x)=2x^4-16x^3 ?
The extreme f(x)=2x^4-16x^3 is Saddle(0,0),Minimum(6,-864)