What is the extreme f(x)=(x^2-7)/(x-4) ?

The extreme f(x)=(x^2-7)/(x-4) is Maximum(1,2),Minimum(7,14)

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extreme f(x)=(x^2-7)/(x-4)

extreme points

f (x) = \frac{x ^{2} - 7}{x - 4}

Maximum (1, 2), Minimum (7, 14)

Solution steps

f^{'} (x) = \frac{x ^{2} - 8 x + 7}{( x - 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 4,

Decreasing

: 4 < x < 7,

Increasing

: 7 < x < \infty

Plug

x = 1

into

\frac{x ^{2} - 7}{x - 4} : 2

Maximum (1, 2)

Plug

x = 7

into

\frac{x ^{2} - 7}{x - 4} : 14

Minimum (7, 14)

Maximum (1, 2), Minimum (7, 14)

\frac{( 3 x - 15 )}{- x ^{2} + 5 x}

- x + 3

f (x) = - \frac{1}{2} x + 15

(- 3, 7), (1, - 1)

f (x) = x^{2} - 3 x

What is the extreme f(x)=(x^2-7)/(x-4) ?
The extreme f(x)=(x^2-7)/(x-4) is Maximum(1,2),Minimum(7,14)