What is the extreme f(x)=(x^2-5x+7)/(x-3) ?

The extreme f(x)=(x^2-5x+7)/(x-3) is Maximum(2,-1),Minimum(4,3)

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extreme f(x)=(x^2-5x+7)/(x-3)

extreme

Maximum (2, - 1), Minimum (4, 3)

Solution steps

f^{'} (x) = \frac{x ^{2} - 6 x + 8}{( x - 3 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 3,

Decreasing

: 3 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 2

into

\frac{x ^{2} - 5 x + 7}{x - 3} : - 1

Maximum (2, - 1)

Plug

x = 4

into

\frac{x ^{2} - 5 x + 7}{x - 3} : 3

Minimum (4, 3)

Maximum (2, - 1), Minimum (4, 3)

What is the extreme f(x)=(x^2-5x+7)/(x-3) ?
The extreme f(x)=(x^2-5x+7)/(x-3) is Maximum(2,-1),Minimum(4,3)