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

The extreme y=(x^2+x-1)/(x-1) is Maximum(0,1),Minimum(2,5)

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

extreme points

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

Maximum (0, 1), Minimum (2, 5)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Decreasing

: 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

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

Maximum (0, 1)

Plug

x = 2

into

\frac{x ^{2} + x - 1}{x - 1} : 5

Minimum (2, 5)

Maximum (0, 1), Minimum (2, 5)

f (x) = \frac{x ^{6}}{10} - \frac{2 x ^{5}}{3}

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

f (x) = 2 x + 3

f (x) = cot (x) + 10, - π \leq x \leq \frac{π}{3}

f (x, y) = x^{2} y + 2 x y - 2 x y^{2} + 2

What is the extreme y=(x^2+x-1)/(x-1) ?
The extreme y=(x^2+x-1)/(x-1) is Maximum(0,1),Minimum(2,5)