Question

Determine the value of ’k’ for which the following function is continuous at x = 3;
f(x) =${ \dfrac{{{{(x + 3)}^2} - 36}}{{x - 3}} ,x \ne 3\\\ \;\;k ,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x = 3 }$

Answer 1

Here you can see the question is based on the continuity of function. You must already know that a function is said to be continuous at a point x=a, in its domain if the following three conditions are satisfied , 1. f(a) exists i.e.: f(a) is finite, 2. exist i.e.: right hand limit is equal to the left hand limit and both are finite. 3. And lastly $\mathop {\lim }\limits_{x \to a} f(x) = f(a)$ . we are given the point where f(x) is continuous and we need to find out the value of k.

Complete step by step solution:
Given data: The given function is

\dfrac{{{{(x + 3)}^2} - 36}}{{x - 3}} ,x \ne 3\\\ k ,x = 3 \right.$$ And the f(x) is continuous at x=3. To find out the value of k. Here, $$f(x) = \dfrac{{{{(x + 3)}^2} - 36}}{{x - 3}}$$ It is given that f(x) is continuous at x=3 and for x=3, f(x) = k . We know that, if a function is a continuous at a point x= a in its domain, it satisfies $$\mathop {\lim }\limits_{x \to a} f(x) = f(a)$$ . For $$f(x) = \dfrac{{{{(x + 3)}^2} - 36}}{{x - 3}}$$ and the point is x . Hence, $ \mathop {\lim }\limits_{x \to 3} f(x) = k\\\ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{{{(x + 3)}^2} - 36}}{{x - 3}}} \right) = k\\\ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{{{(x + 3)}^2} - {6^2}}}{{x - 3}}} \right) = k .....(i) $ You can see, $${(x + 3)^2} - {6^2}$$ is in the form of $${A^2} - {B^{2\,}}\,where\,\,A = x + 3\,\,and\,\,B = 6\,$$ . We knew that , $${A^2} - {B^{2\,}} = (A - B)(A + B)$$ So, equation (i) can be written as, $ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{(x + 3 + 6)(x + 3 - 6}}{{x - 3}}} \right) = k\\\ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{(x + 9)(x - 3)}}{{x - 3}}} \right) = k\\\ \Rightarrow \mathop {\lim }\limits_{x \to 3} (x + 9) = k .....(ii) $ Since, $$x \to 3,\,\,\mathop {\lim }\limits_{x \to 3} (x + a)$$ can be written as $ \Rightarrow \mathop {\lim }\limits_{x \to 3} (x + 9) = 3 + 9\\\ \Rightarrow k = 12 [Using\,\,(ii)] $ **Hence, the required value of k is 12.** **Note:** Students often make mistakes in understanding the continuity of function where it is continuous and where it is not. Always keep in mind the conditions that are satisfied where a continuous at a certain point. Try to withdraw the limit after making the function as simple as possible. So that it avoids mistakes in calculation.