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

The extreme y=(x^2-2x+1)/(x^2+3) is Maximum(-3, 4/3),Minimum(1,0)

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

extreme points

y = \frac{x ^{2} - 2 x + 1}{x ^{2} + 3}

Maximum (- 3, \frac{4}{3}), Minimum (1, 0)

Solution steps

f^{'} (x) = \frac{2 x ^{2} + 4 x - 6}{( x ^{2} + 3 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 3

into

\frac{x ^{2} - 2 x + 1}{x ^{2} + 3} : \frac{4}{3}

Maximum (- 3, \frac{4}{3})

Plug

x = 1

into

\frac{x ^{2} - 2 x + 1}{x ^{2} + 3} : 0

Minimum (1, 0)

Maximum (- 3, \frac{4}{3}), Minimum (1, 0)

f (x) = (7 + x) (11 - 3 x)^{\frac{1}{3}}

y = x^{- 4} \cdot e^{x^{4}}

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

f (x) = - 2 x^{2} + 3 x - 4

f (x) = \frac{2 + 4 x ^{2}}{1 - 3 x}

What is the extreme y=(x^2-2x+1)/(x^2+3) ?
The extreme y=(x^2-2x+1)/(x^2+3) is Maximum(-3, 4/3),Minimum(1,0)