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

The extreme f(x)=(6x^3)/(x-4) is Saddle(0,0),Minimum(6,648)

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extreme f(x)=(6x^3)/(x-4)

extreme points

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

Saddle (0, 0), Minimum (6, 648)

Solution steps

f^{'} (x) = \frac{6 ( 2 x ^{3} - 12 x ^{2} )}{( x - 4 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Decreasing

: 4 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 0

into

\frac{6 x ^{3}}{x - 4} : 0

Saddle (0, 0)

Plug

x = 6

into

\frac{6 x ^{3}}{x - 4} : 648

Minimum (6, 648)

Saddle (0, 0), Minimum (6, 648)

f (x) = 2 x^{5} + 10 x^{4} + x^{3} - 10

f (4) = e^{2 x^{2} + 4 y^{2} - 16 x}

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

y = 5 cos^{2} (x)

h (x, t) = 1.5 + lo g_{3} (x) (t + 1)

What is the extreme f(x)=(6x^3)/(x-4) ?
The extreme f(x)=(6x^3)/(x-4) is Saddle(0,0),Minimum(6,648)