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

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

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

extreme points

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

Maximum (- 1, - 2), Minimum (3, 6)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 1,

Decreasing

: 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 1

into

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

Maximum (- 1, - 2)

Plug

x = 3

into

\frac{x ^{2} + 3}{1 x - 1} : 6

Minimum (3, 6)

Maximum (- 1, - 2), Minimum (3, 6)

f (x) = x^{\frac{2}{3}} (x - 7)

x^{4} + x^{2} - 10 x + 25

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

f (x, y) = 60 x + 34 y - 4 x y - 6 x^{2} - 3 y^{2} + 5

f (x) = x^{2} \cdot e^{- x^{2}}

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