What is the extreme f(x)=((x^2-12))/(x-4) ?

The extreme f(x)=((x^2-12))/(x-4) is Maximum(2,4),Minimum(6,12)

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extreme f(x)=((x^2-12))/(x-4)

extreme points

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

Maximum (2, 4), Minimum (6, 12)

Solution steps

f^{'} (x) = \frac{x ^{2} - 8 x + 12}{( x - 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 4,

Decreasing

: 4 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 2

into

\frac{x ^{2} - 12}{x - 4} : 4

Maximum (2, 4)

Plug

x = 6

into

\frac{x ^{2} - 12}{x - 4} : 12

Minimum (6, 12)

Maximum (2, 4), Minimum (6, 12)

f (t) = t e^{\frac{w}{t}}

f (x) = y^{2} - x y - x^{2}

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

f (x, y) = \frac{( 2 x + y )}{3}

f (x, y) = \frac{x - x ^{2}}{1 - y ^{2}}

What is the extreme f(x)=((x^2-12))/(x-4) ?
The extreme f(x)=((x^2-12))/(x-4) is Maximum(2,4),Minimum(6,12)