What is the extreme ((x^2-1))/((x-3)^2) ?

The extreme ((x^2-1))/((x-3)^2) is Minimum(1/3 ,-1/8)

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extreme ((x^2-1))/((x-3)^2)

extreme points

\frac{( x ^{2} - 1 )}{( x - 3 ) ^{2}}

Minimum (\frac{1}{3}, - \frac{1}{8})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < \frac{1}{3},

Increasing

: \frac{1}{3} < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = \frac{1}{3}

into

\frac{x ^{2} - 1}{( x - 3 ) ^{2}} : - \frac{1}{8}

Minimum (\frac{1}{3}, - \frac{1}{8})

s (t) = - 4.9 t^{2} + 29 t + 2

y = (x + 3)^{2} - 7

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

f (x) = 20.85 x y + 10.7 x^{2} + 51.75 y^{2}

x^{5} - x^{3}

What is the extreme ((x^2-1))/((x-3)^2) ?
The extreme ((x^2-1))/((x-3)^2) is Minimum(1/3 ,-1/8)