Question

The function \[f(x) = \dfrac{{x(x - 2)}}{{x - 1}}\]is continuous at \[x = 1\]
a. true
b. false

Answer 1

Hint: Firstly we try to check if the function is well defined at the point \[x = 1\]. If it is not well defined then there is no question of it to be continuous at that point. Otherwise we need to check if \[{\lim _{x \to a}}f(x) = f(a)\]as it is the condition of a function to be continuous at a given point a.

Complete step by step solution: A function is continuous at a point a if \[{\lim _{x \to a}}f(x) = f(a)\].
Now it is given, \[f(x) = \dfrac{{x(x - 2)}}{{x - 1}}\]and also given, \[x = 1\]
So, \[a = 1\], we have, \[f(a) = f(1) = \]\[\dfrac{{1(1 - 2)}}{{1 - 1}}\]which is not well defined, as denominator will be 0.
Now, for a function to be continuous at a point it should be well defined and should exist with respect to that point.
Here we can easily see, that the function is not properly defined at the given point. So, the function is not continuous at a given point.
So, the given statement is false.

Hence, option (b) is the correct.

Note: In this problem, we are using the continuity of the function,

Here are some of the properties of the function of continuity,
1) If a function is continuous at a point an if \[{\lim _{x \to a}}f(x) = f(a)\].
2) The sum, difference, and product of two continuous functions are each continuous functions. All polynomial functions are continuous.
3) The quotient of two continuous functions is continuous where it is defined. (It won't be defined when the denominator is zero.) All rational functions (quotients of two polynomials) are continuous where they're defined.
4) The composition of two continuous functions is continuous. So, for example, the square root function is continuous, so the square root of a continuous function is another continuous function.