What is the extreme f(x)=(y-3)/(y^2-3y+9) ?

The extreme f(x)=(y-3)/(y^2-3y+9) is Minimum(0,-1/3),Maximum(6, 1/9)

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extreme f(x)=(y-3)/(y^2-3y+9)

extreme points

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

Minimum (0, - \frac{1}{3}), Maximum (6, \frac{1}{9})

Solution steps

f^{'} (y) = \frac{- y ^{2} + 6 y}{( y ^{2} - 3 y + 9 ) ^{2}}

Find intervals:

Decreasing

: - \infty < y < 0,

Increasing

: 0 < y < 6,

Decreasing

: 6 < y < \infty

Plug

y = 0

into

\frac{y - 3}{y ^{2} - 3 y + 9} : - \frac{1}{3}

Minimum (0, - \frac{1}{3})

Plug

y = 6

into

\frac{y - 3}{y ^{2} - 3 y + 9} : \frac{1}{9}

Maximum (6, \frac{1}{9})

Minimum (0, - \frac{1}{3}), Maximum (6, \frac{1}{9})

f (x) = (16 - x) (100 x + 1300)

f (x) = \frac{1}{3 x ^{\frac{2}{3}}}

\frac{2 x - 3}{x}

f (x) = x^{3} - x^{2} - x + 6, - 1 \leq x \leq 2

f (x) = 3 x^{5} - 17 x^{3} + 12 x^{2} - 3 x + π

What is the extreme f(x)=(y-3)/(y^2-3y+9) ?
The extreme f(x)=(y-3)/(y^2-3y+9) is Minimum(0,-1/3),Maximum(6, 1/9)