# find the critical numbers of f(x)=(x^2)/(x-1)?

the answer says its only x=0,2, why isn't 1 part of the answer since f' DNE at x=1

### 5 Answers

- TomVLv 71 month agoFavorite Answer
(1, f(1)) is not a critical point because the definition of a critical point requires that the point be within the domain of the function.

x = 1 is not in the domain of f(x) = x²/(x-1). Therefore f(1) is undefined and cannot be a critical point of f(x).

- Wayne DeguManLv 71 month ago
The technical definition of a critical point is where f '(x) = 0

so, we could have local minimum, maximum or even points of inflection.

The rational function does not exist for x = 1, as you state. In terms of analysing and sketching the function, x = 1 can be thought of as a critical value as x = 1 is a vertical asymptote.

Using the quotient rule we have that:

f '(x) = [2x(x - 1) - (1)(x²)]/(x - 1)²

so, f '(x) = x(x - 2)/(x - 1)²

Hence, f '(x) = 0 when x = 0 and when x = 2...which are critical points.

It can be shown that (0, 0) is a local maximum and (2, 4) is a local minimum.

A sketch is below.

:)>

- PhilipLv 61 month ago
f(x) = (x^2)/(x-1) = (x^2)(x-1)^(-1);

f'(x) = 2x(x-1)^(-1) -(x^2)(x-1)^(-2). (1/x)f'(x)(x-1)^2 = 2(x-1) -x = x-2, ie.,

f'(x) = x(x-2)/(x-1)^2.;

By definition, critical numbers of f(x) occur where f'(x) = 0. Not where f'(x) does

not exist.

- micatkieLv 71 month ago
f(x) = x²/(x - 1)

f'(x) = (d/dx)[x²/(x - 1)]

f'(x) = [(x - 1)(x²)' - x²(x - 1)'] / (x - 1)²

f'(x) = [(x - 1)(2x) - x²(1)] / (x - 1)²

f'(x) = [2x² - 2x - x²] / (x - 1)²

f'(x) = (x² - 2x) / (x - 1)²

f'(x) = x(x - 2) / (x - 1)²

When f'(x) = 0:

x = 0 or x = 2

Hence, f(x) has critical points at x = 0, 2

There is NO critical point at x = 1.

Actually, f(x) = 1/0 at x = 1

In other words, f(x) is undefined at x = 1

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- King LeoLv 71 month ago
In a normal world x = 1 should be an answer as per definition of a critical point in calculus