 # Find the constant a such that the function is continuous on the entire real line.

Given Function: $\ f\left( x\right)= \bigg\{\begin{array}{rcl} x^3, & x≤2 \\ ax^2, & x>2 \end{array}$

The aim of the question is to find the value of constant a for which the given function will be continuous on the whole real number line.

The basic concept behind this question is the knowledge of the Continuous Function.

Given function in the question is:

$\ f\left( x\right)= \bigg\{ \begin{array}{rcl} x3, & x≤2 \\ ax^2, & x>2 \end{array}$

We know that if $f$ is a continuous function then, then it will be also continuous at $x=2$.

$\lim_ { x \rightarrow 2^{+}}\ \ {f\left(x\right)\ }=\ \lim_{x\rightarrow2^-}\ \ {f\left(x\right)\ }=\lim_{x\rightarrow2}\ \ {f\left(x\right)\ }=\ {f\left(2\right)\ }$

$\lim_{x\rightarrow2^+}\ \ {f\left(x\right)\ }=\ ax^2$

Given we know that $x>2$ so putting to see if the function is continuous at $x=2$ put the value of $x$ here equal to $2$.

$\lim_{x\rightarrow2^+}\ \ {f\left(x\right)\ }=\ a{(2)}^2$

$\lim_{x\rightarrow2^+}\ \ {f\left(x\right)\ }=\ 4a$

Now for the other equation we have:

$\lim_{x\rightarrow2^-}\ \ {f\left(x\right)\ }=\ x^3$

Given we know that $x\le2$ so putting to see if the function is continuous at $x=2$ put the value of $x$ here equal to $2$.

$\lim_{x\rightarrow2^-}\ \ {f\left(x\right)\ }=\ {(2)}^3$

$\lim_{x\rightarrow2^-}\ \ {f\left(x\right)\ }=\ 8$

From the above equations, we know that:

$\lim_{x\rightarrow2^+}\ \ {f\left(x\right)\ }=\ \lim_{x\rightarrow2^-}\ \ {f\left(x\right)\ }$

Putting values of both limits here, we get:

$\lim_{x\rightarrow2^+}\ \ {f\left(x\right)\ }=\ 4a$

And:

$\lim_{x\rightarrow2^-}\ \ {f\left(x\right)\ }=\ 8$

$4a = 8$

From above equation we find out the value of $a$:

$a = \frac {8}{4 }$

$a = 2$

So the value of constant $a$ is $2$ for which the given function $\ f\left( x\right)= \bigg\{ \begin{array}{rcl} x3, & x≤2 \\ ax^2, & x>2 \end{array}$  is continuous on the whole real number line.

## Numerical Result

$\lim_{x\rightarrow 2^+}\ \ {f\left(x\right)\ }=\ \lim_{x\rightarrow2^-}\ \ {f\left(x\right)\ }$

Values of both the limits are:

$\lim_{x \rightarrow 2^+}\ \ {f\left(x\right)\ }=\ 4a$

$\lim_{x\rightarrow 2^-}\ \ {f\left(x\right)\ }=\ 8$

Putting it in the above equation, we get the following equation:

$4a =8$

From the above equation, we can easily find out the value of $a$:

$a = \frac {8}{4 }$

$a = 2$

## Example

Find out the value of constant $a$ for the function:

$\ f\left( x\right)= \bigg\{ \begin{array}{rcl} x3, & x≤4 \\ ax^2, & x>4 \end{array}$

Solution

We know that if $f$ is a continuous function, then it will be also continuous at $x=4$.

$\lim_ { x \rightarrow 4^{+}}\ \ {f\left(x\right)\ }=\ \lim_{x\rightarrow4^-}\ \ {f\left(x\right)\ }=\lim_{x\rightarrow4}\ \ {f\left(x\right)\ }=\ {f\left(4\right)\ }$

$\lim_{x\rightarrow4^+}\ \ {f\left(x\right)\ }=\ ax^2$

$\lim_{x\rightarrow4^+}\ \ {f\left(x\right)\ }=\ a{(4)}^2$

$\lim_{x\rightarrow4^+}\ \ {f\left(x\right)\ }=\ 16a$

$\lim_{x\rightarrow4^-}\ \ {f\left(x\right)\ }=\ x^3$

$\lim_{x\rightarrow4^-}\ \ {f\left(x\right)\ }=\ {(4)}^3$

$\lim_{x\rightarrow4^-}\ \ {f\left(x\right)\ }=\ 64$

Equating both equations:

$16a=64$

$a=\frac {64}{16}$

$a=4$