What is the extreme f(x)=(12)/(6-x)-x/3+8 ?

The extreme f(x)=(12)/(6-x)-x/3+8 is Minimum(0,10),Maximum(12,2)

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extreme f(x)=(12)/(6-x)-x/3+8

extreme points

f (x) = \frac{12}{6 - x} - \frac{x}{3} + 8

Minimum (0, 10), Maximum (12, 2)

Solution steps

f^{'} (x) = \frac{12}{( - x + 6 ) ^{2}} - \frac{1}{3}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 6,

Increasing

: 6 < x < 12,

Decreasing

: 12 < x < \infty

Plug

x = 0

into

\frac{12}{6 - x} - \frac{x}{3} + 8 : 10

Minimum (0, 10)

Plug

x = 12

into

\frac{12}{6 - x} - \frac{x}{3} + 8 : 2

Maximum (12, 2)

Minimum (0, 10), Maximum (12, 2)

y = - 2 x^{2} + 12 x - 8

f (x) = x^{3} - 3 x^{2} - 9 x + 6

f (x, y) = ln (4 x^{2} + 9 y^{2} - 36)

x^{3} - x^{2} - x + 2

x^{3} - x^{2} - x + 3

What is the extreme f(x)=(12)/(6-x)-x/3+8 ?
The extreme f(x)=(12)/(6-x)-x/3+8 is Minimum(0,10),Maximum(12,2)