Question

Find a and b if the function
$f\left( x \right)=\left\{ \begin{matrix}
   \dfrac{\sin x}{x}\text{ }-2\le x<0 \\
   a{{.2}^{x}}\text{ 0}\le \text{x}\le \text{1} \\
   b+x\text{ 1x}\le \text{2} \\
\end{matrix} \right.$
is continuous in the interval $\left[ -2,2 \right]$

Answer 1

Hint: To solve this problem, we should know the concept of continuity of the functions. A function $f\left( x \right)$ is continuous at a point $x={{x}_{1}}$ if and only if the function’s limit from the left and from the right of the value $x={{x}_{1}}$ and the left limit and the right limit are equal to the functional value at $x={{x}_{1}}$. Mathematically, it can be written as
Left limit = Right limit = $f\left( {{x}_{1}} \right)$

Complete answer:
$\underset{x\to x_{1}^{-}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to x_{1}^{+}}{\mathop{\lim }}\,f\left( x \right)=f\left( {{x}_{1}} \right)$
In some cases, the function itself will not be defined to the left or to the right of the value. Then we consider the limit value to be equal to the functional value for the function to be continuous at that value.
Using this property, we can write that the function $\dfrac{\sin x}{x}$ is the left functional definition at x = 0 and $a{{.2}^{x}}$ is the right functional definition at x = 0. We can infer from the question that the function should be continuous in the whole range of $\left[ -2,2 \right]$, we can apply the continuous condition for the given function at x = 0. We will get the value of a from this calculation. Likewise, we can also calculate the value of b by applying the continuous condition at x = 1.

We know that the condition for a function to be continuous at a given point is that
A function $f\left( x \right)$ is continuous at a point $x={{x}_{1}}$ if and only if the function’s limit from the left and from the right of the value $x={{x}_{1}}$ and the left limit and the right limit are equal to the functional value at $x={{x}_{1}}$. Mathematically, it can be written as
Left limit = Right limit = $f\left( {{x}_{1}} \right)$
$\underset{x\to x_{1}^{-}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to x_{1}^{+}}{\mathop{\lim }}\,f\left( x \right)=f\left( {{x}_{1}} \right)$

Consider the function at the point x = 0
The left functional value is $\dfrac{\sin x}{x}$ and applying the limit, we get
Left limit = \[\underset{x\to 0_{{}}^{-}}{\mathop{\lim }}\,\dfrac{\sin x}{x}\]
We know that $x\approx 0\Rightarrow \sin x\approx x$
Using this relation, we get
Left limit = \[\underset{x\to 0_{{}}^{-}}{\mathop{\lim }}\,\dfrac{x}{x}=\underset{x\to 0_{{}}^{-}}{\mathop{\lim }}\,1=1\]
The right functional value is $a{{.2}^{x}}$ and applying the limit, we get
Right limit = \[\underset{x\to 0_{{}}^{+}}{\mathop{\lim }}\,a{{.2}^{x}}=\underset{x\to 0_{{}}^{+}}{\mathop{\lim }}\,a{{.2}^{0}}=\underset{x\to 0_{{}}^{+}}{\mathop{\lim }}\,a=a\]
The functional value at x = 0 is $f\left( 0 \right)=a{{.2}^{0}}=a$
Using the relation Left limit = Right limit = $f\left( {{x}_{1}} \right)$, we get
$\begin{align}
  & 1=a=a \\
 & a=1 \\
\end{align}$
Consider the function at the point x = 1
The left functional value is $a{{.2}^{x}}={{2}^{x}}$ and applying the limit, we get
Left limit = \[\underset{x\to 1_{{}}^{-}}{\mathop{\lim }}\,{{2}^{x}}=2\]
The right functional value is $b+x$ and applying the limit, we get
Right limit = \[\underset{x\to 0_{{}}^{+}}{\mathop{\lim }}\,b+x=\underset{x\to 0_{{}}^{+}}{\mathop{\lim }}\,b+1=b+1\]
The functional value at x = 1 is $f\left( 1 \right)={{2}^{1}}=2$
Using the relation Left limit = Right limit = $f\left( {{x}_{1}} \right)$, we get
$\begin{align}
  & 2=b+1=2 \\
 & b=1 \\
\end{align}$
$\therefore $ The values of a and b are 1 and 1 respectively.

Note: In the question, we are given the condition of continuity between $\left[ -2,2 \right]$ but we just checked the continuity at x = 0, 1. This is because, 0 and 1 are the transition points at which the function is changing its nature. So, they are the main points. We should also note that the functions like ${{2}^{x}}$ and $x+b$ are always continuous within the range they are specified. But the functions like $\dfrac{1}{x}$, $\dfrac{1}{\sin x}$ are not continuous at some points. We should be aware of the range in which the function is defined. If there is a point within the range where there is no transition but the function is not defined, it is also a discontinuous point.