Question

If $f(x) = \left\{
  \dfrac{{{2^{x + 2}} - 16}}{{{4^x} - 16}},{\text{ if }}x \ne 2 \\
  k,{\text{ if }}x = 2 \\
  \right.$ is continuous at x = 2 , find k.

Answer 1

Hint:
As the given function is continuous we know that for every continuous function $\mathop {\lim }\limits_{x \to a} f(x) = f(a)$ using this and substituting 2 instead of a we get the required value of k.

Complete step by step solution:
We are given that the function f(x) is continuous at x=2
Whenever a given function is continuous at x = a then
$ \Rightarrow \mathop {\lim }\limits_{x \to a} f(x) = f(a)$
And here are given function is continuous at 2
Hence
$ \Rightarrow \mathop {\lim }\limits_{x \to 2} f(x) = f(2)$
Hence from the given function we get
$ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{{2^{x + 2}} - 16}}{{{4^x} - 16}} = k$
Here ${2^{x + 2}}$can be written as ${2^x}{.2^2}$
 $
   \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{{2^x}{{.2}^2} - 16}}{{{4^x} - 16}} = k \\
   \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{{{4.2}^x} - 16}}{{{4^x} - 16}} = k \\
   \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{4\left( {{2^x} - 4} \right)}}{{{{\left( {{2^x}} \right)}^2} - 16}} = k \\
 $
The denominator can be expanded using the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
$
   \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{4\left( {{2^x} - 4} \right)}}{{\left( {{2^x} + 4} \right)\left( {{2^x} - 4} \right)}} = k \\
   \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{4}{{\left( {{2^x} + 4} \right)}} = k \\
 $
Now as the limit tends to 2 lets substitute 2 in the place of x
$
   \Rightarrow \dfrac{4}{{\left( {{2^2} + 4} \right)}} = k \\
   \Rightarrow \dfrac{4}{{4 + 4}} = k \\
   \Rightarrow \dfrac{4}{8} = k \\
   \Rightarrow \dfrac{1}{2} = k \\
 $
Hence we get the value of k.

The value of k is $\dfrac{1}{2}$

Note:
A mathematical function is called continuous if, roughly said, a small change in the input only causes a small change in the output. If this is not the case, the function is discontinuous. Functions defined on the real numbers, with one input and one output variable, will show as an uninterrupted line (or curve). They can be drawn without lifting the pen off of the page.