What is the extreme f(x)=(x^2-16x+64)/(x-10) ?

The extreme f(x)=(x^2-16x+64)/(x-10) is Maximum(8,0),Minimum(12,8)

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extreme f(x)=(x^2-16x+64)/(x-10)

extreme

Maximum (8, 0), Minimum (12, 8)

Solution steps

f^{'} (x) = \frac{x ^{2} - 20 x + 96}{( x - 10 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 8,

Decreasing

: 8 < x < 10,

Decreasing

: 10 < x < 12,

Increasing

: 12 < x < \infty

Plug

x = 8

into

\frac{x ^{2} - 16 x + 64}{x - 10} : 0

Maximum (8, 0)

Plug

x = 12

into

\frac{x ^{2} - 16 x + 64}{x - 10} : 8

Minimum (12, 8)

Maximum (8, 0), Minimum (12, 8)

What is the extreme f(x)=(x^2-16x+64)/(x-10) ?
The extreme f(x)=(x^2-16x+64)/(x-10) is Maximum(8,0),Minimum(12,8)