What is the extreme f(x)=(x^2-15)/(x-4) ?

The extreme f(x)=(x^2-15)/(x-4) is Maximum(3,6),Minimum(5,10)

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extreme f(x)=(x^2-15)/(x-4)

extreme points

f (x) = \frac{x ^{2} - 15}{x - 4}

Maximum (3, 6), Minimum (5, 10)

Solution steps

f^{'} (x) = \frac{x ^{2} - 8 x + 15}{( x - 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 3,

Decreasing

: 3 < x < 4,

Decreasing

: 4 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = 3

into

\frac{x ^{2} - 15}{x - 4} : 6

Maximum (3, 6)

Plug

x = 5

into

\frac{x ^{2} - 15}{x - 4} : 10

Minimum (5, 10)

Maximum (3, 6), Minimum (5, 10)

f (x) = x + 5 - x

f (x) = ln (x - 3)

\frac{3 x ^{4} + 1}{x ^{3}}

\frac{e ^{x}}{8 + e ^{x}}

f (x) = \frac{5 x - 6}{2 x + 1}

What is the extreme f(x)=(x^2-15)/(x-4) ?
The extreme f(x)=(x^2-15)/(x-4) is Maximum(3,6),Minimum(5,10)