What is the extreme f(x)=(x^2-10x+21)/(2x-15) ?

The extreme f(x)=(x^2-10x+21)/(2x-15) is Maximum(6,1),Minimum(9,4)

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extreme f(x)=(x^2-10x+21)/(2x-15)

extreme points

f (x) = \frac{x ^{2} - 10 x + 21}{2 x - 15}

Maximum (6, 1), Minimum (9, 4)

Solution steps

f^{'} (x) = \frac{2 x ^{2} - 30 x + 108}{( 2 x - 15 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 6,

Decreasing

: 6 < x < \frac{15}{2},

Decreasing

: \frac{15}{2} < x < 9,

Increasing

: 9 < x < \infty

Plug

x = 6

into

\frac{x ^{2} - 10 x + 21}{2 x - 15} : 1

Maximum (6, 1)

Plug

x = 9

into

\frac{x ^{2} - 10 x + 21}{2 x - 15} : 4

Minimum (9, 4)

Maximum (6, 1), Minimum (9, 4)

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

f (x, y) = x^{2} - 6 x y + y^{3} - 3

f (x) = (x + 9)^{\frac{2}{3}}

- x^{2} + 10 x - 21

f (x) = \frac{x ^{2}}{x ^{2} + 243}

What is the extreme f(x)=(x^2-10x+21)/(2x-15) ?
The extreme f(x)=(x^2-10x+21)/(2x-15) is Maximum(6,1),Minimum(9,4)