What is the extreme f(x)=4x^4-32x^3 ?

The extreme f(x)=4x^4-32x^3 is Saddle(0,0),Minimum(6,-1728)

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extreme f(x)=4x^4-32x^3

extreme points

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

Saddle (0, 0), Minimum (6, - 1728)

Solution steps

f^{'} (x) = 16 x^{3} - 96 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 0

into

4 x^{4} - 32 x^{3} : 0

Saddle (0, 0)

Plug

x = 6

into

4 x^{4} - 32 x^{3} : - 1728

Minimum (6, - 1728)

Saddle (0, 0), Minimum (6, - 1728)

y = 3 x^{\frac{2}{3}} - 2 x [- 1.1]

f (x) = \frac{7 + x}{8 - x}

f (x) = x^{7} - x^{5}

f (x) = - (5 x + 3) e^{- 2 x}, (- 3, 2)

x x - 1

What is the extreme f(x)=4x^4-32x^3 ?
The extreme f(x)=4x^4-32x^3 is Saddle(0,0),Minimum(6,-1728)