What is the extreme f(x)=(x-5)/(x^2-6x+9) ?

The extreme f(x)=(x-5)/(x^2-6x+9) is Maximum(7, 1/8)

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extreme f(x)=(x-5)/(x^2-6x+9)

extreme points

f (x) = \frac{x - 5}{x ^{2} - 6 x + 9}

Maximum (7, \frac{1}{8})

Solution steps

f^{'} (x) = - \frac{x - 7}{( x - 3 ) ^{3}}

Find intervals:

Decreasing

: - \infty < x < 3,

Increasing

: 3 < x < 7,

Decreasing

: 7 < x < \infty

Plug

x = 7

into

\frac{x - 5}{x ^{2} - 6 x + 9} : \frac{1}{8}

Maximum (7, \frac{1}{8})

y = x^{3} - 12 x - 5

y = x^{3} - 12 x + 8

\frac{x ^{2}}{x + 3}

f (x) = sin (4 x), 0 \leq x \leq \frac{π}{2}

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

What is the extreme f(x)=(x-5)/(x^2-6x+9) ?
The extreme f(x)=(x-5)/(x^2-6x+9) is Maximum(7, 1/8)