What is the extreme (x^2-48)/(x-7) ?

The extreme (x^2-48)/(x-7) is Maximum(6,12),Minimum(8,16)

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extreme (x^2-48)/(x-7)

extreme points

\frac{x ^{2} - 48}{x - 7}

Maximum (6, 12), Minimum (8, 16)

Solution steps

f^{'} (x) = \frac{x ^{2} - 14 x + 48}{( x - 7 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 6,

Decreasing

: 6 < x < 7,

Decreasing

: 7 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 6

into

\frac{x ^{2} - 48}{x - 7} : 12

Maximum (6, 12)

Plug

x = 8

into

\frac{x ^{2} - 48}{x - 7} : 16

Minimum (8, 16)

Maximum (6, 12), Minimum (8, 16)

What is the extreme (x^2-48)/(x-7) ?
The extreme (x^2-48)/(x-7) is Maximum(6,12),Minimum(8,16)