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

The extreme f(x)=(x+3)/(x^2-2x+1) is Minimum(-7,-1/16)

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extreme f(x)=(x+3)/(x^2-2x+1)

extreme points

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

Minimum (- 7, - \frac{1}{16})

Solution steps

f^{'} (x) = - \frac{x + 7}{( x - 1 ) ^{3}}

Find intervals:

Decreasing

: - \infty < x < - 7,

Increasing

: - 7 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 7

into

\frac{x + 3}{x ^{2} - 2 x + 1} : - \frac{1}{16}

Minimum (- 7, - \frac{1}{16})

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

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

x^{2} - 4 x + 1

f (x) = x \frac{4}{3} (x - 4)

f (x) = x^{4} - 8 x^{2} - 6

What is the extreme f(x)=(x+3)/(x^2-2x+1) ?
The extreme f(x)=(x+3)/(x^2-2x+1) is Minimum(-7,-1/16)