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

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)

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)