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 points

f (x) = \frac{x ^{2} - 5 x + 7}{x - 3}

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)

f (x) = (3 x^{3} - 9)

f (x) = \frac{x ^{2} - 4}{e ^{x^{4} - 1}}

f (x) = 2 x - 5 x^{2} + 4

f (x) = \frac{4 x}{x ^{2} + 16}

f (x, y) = x^{4} + y^{4} - 4 x y + x

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)