Continuity of piecewise functions

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Here we use limits to check whether piecewise functions are continuous.

Before we start talking about continuity of piecewise functions, let’s remind ourselves of our parent functions that are continuous on their domains.

Continuity of Parent Functions The following functions are continuous on their natural domains:

Constant functions
Power functions
Polynomials
Rational functions
Exponential functions
Logarithmic functions

In essence, we are saying that the functions listed above are continuous wherever they are defined.

We can prove continuity of polynomials earlier using the Sum Law, Product Law and continuity of power functions.

We can prove continuity of rational functions earlier using the Quotient Law and continuity of polynomials.

Since a continuous function and its inverse have “unbroken” graphs, it follows that an inverse of a continuous function is continuous on its domain.

Using the Limit Laws we can prove that given two functions, both continuous on the same interval, then their sum, difference, product, and quotient (where defined) are also continuous on the same interval (where defined).

In this section we will work a couple of examples involving limits, continuity and piecewise functions.

Consider the following piecewise defined function $f(x) = \begin {cases} \frac {x}{x-1} &\text {if $x<0$,}\\ e^{-x} + c &\text {if $x\ge 0$}. \end {cases}$ Find the constant $c$ so that $f$ is continuous at $x=0$ .

To find $c$ such that $f$ is continuous at $x=0$ , we need to find $c$ such that $\lim _{x\to 0} f(x) = f\Bigl (\answer [given]{0}\Bigr ).$ In this case, in order to compute the limit, we will have to compute two one-sided limits, since the expression for $f(x)$ if $x<0$ is different from the expression for $f(x)$ if $x>0$ . So, $\begin{align*} \lim_{x\to 0^-} f(x) &= \lim_{x\to 0^-}\answer[given]{\frac{x}{x-1}}\\ &= \frac{\answer[given]{0}}{-1}\\ &=\answer[given]{0}. \end{align*}$

and

$\begin{align*} \lim_{x\to 0^+} f(x) &= \lim_{x\to 0^+}\left(e^{-x}+c\right)\\ &= e^{\answer[given]{0}} + c\\ &= \answer[given]{1+ c} \end{align*}$

In order for $f$ to be continuous, the limit $\lim _{x\to 0} f(x)$ has to exist. This means that $1 + c = 0\qquad \text {so}\qquad c = \answer [given]{-1}.$ So, if $c=-1$ , the limit $\lim _{x\to 0} f(x)=0$ . Now, we have find the value

$f(0) = \answer [given]{0}.$ Therefore, $\lim _{x\to 0} f(x) = f\Bigl (0\Bigr )\checkmark ,$ which proves that $f$ is continuous at $x=0$ .

Consider the next, more challenging example.

Consider the following piecewise defined function $f(x) = \begin {cases} x+4 &\text {if $x<1$,}\\ ax^2+bx+2 &\text {if $1\le x< 3$,}\\ 6x+a-b &\text {if $x\ge 3$.} \end {cases}$ Find the constants $a$ and $b$ so that $f$ is continuous at both $x=1$ and $x=3$ .

This problem is more challenging because we have more unknowns. However, be brave intrepid mathematician. To find $a$ and $b$ that make $f$ continuous at $x=1$ , we need to find $a$ and $b$ such that $\lim _{x\to 1} f(x) =f(\answer [given]{1}).$ Since $f(1)=a+b+2$ , it follows that $\lim _{x\to 1} f(x) =a+b+2.$ We have to compute two one-sided limits, since the expression for $f(x)$ if $x<1$ is different from the expression for $f(x)$ if $x>1$ . Looking at the limit from the left, we have $\begin{align*} \lim_{x\to 1^-} f(x) &= \lim_{x\to 1^-} \left(\answer[given]{x+4}\right) \\ &=\answer[given]{5}. \end{align*}$

Looking at the limit from the right, we have

$\begin{align*} \lim_{x\to 1^+} f(x) &= \lim_{x\to 1^+} \left(ax^2+bx+2\right) \\ &= \answer[given]{a+b+2}. \end{align*}$

Hence, for the limit $\lim _{x\to 1} f(x)$ to exist, we must have that

$\begin{align*} 5 &= a+b+2\\ or\\ a+b &= \answer[given]{3}. \end{align*}$

Hmmmm. We can’t solve for $a$ or $b$ yet, because we only have this one equality. More work needs to be done.

But we have another point to work with. Specifically we also want to find $a$ and $b$ that make $f$ continuous at $x=3$ . This means we need to find $a$ and $b$ such that $\lim _{x\to 3} f(x) =f(3).$ Since $f(3)=18+a-b$ , it follows that $\lim _{x\to 3} f(x) =18+a-b.$ Looking at the limit from the left, we have

$\begin{align*} \lim_{x\to 3^-} f(x) &= \lim_{x\to 3^-} \left(ax^2+bx+2\right) \\ &=a\cdot 9 + b\cdot 3 + 2. \end{align*}$

Looking at the limit from the right, we have

$\begin{align*} \lim_{x\to 3^+} f(x) &= \lim_{x\to 3^+} \left(6x+a-b\right) \\ &= \answer[given]{18+a-b}. \end{align*}$

Hence

$\begin{align*} 9a + 3b + 2 &= 18+a-b\\ 8a + 4b -16 &= 0\\ 2a + b -4 &= 0 \end{align*}$

So now we have two equations and two unknowns: $a+b=3 \qquad \text {and}\qquad 2a + b = 4.$ Set $b = 3-a$ and write

$\begin{align*} 2a+3-a&=4 \\ a &= 1, \end{align*}$

hence $a=\answer [given]{1}\qquad \text {and so}\qquad b =\answer [given]{2}.$ Let’s check by plugging in the values for both $a$ and $b$ ; we find $f(x) = \begin {cases} x+4 &\text {if $x<1$,}\\ \answer [given]{x^2+2x+2} &\text {if $1\le x< 3$,}\\ \answer [given]{6x-1} &\text {if $x\ge 3$.} \end {cases}$ Now $\lim _{x\to 1^-} f(x) =\lim _{x\to 1^+}f(x) =f(1) = 5,$ and $\lim _{x\to 3^-} f(x) =\lim _{x\to 3^+}f(x) =f(3) = 17.$ So setting $a= \answer [given]{1}$ and $b=\answer [given]{2}$ makes $f$ continuous at $x=1$ and $x=3$ .

We can confirm our results by looking at the graph of $y=f(x)$ : $\graph [xmin=-10,xmax=10,ymin=0,ymax=30]{y=x+4\left \{x<1\right \},y=x^2+2x+2\left \{1\leq x<3\right \},y=6x-1\left \{3 \leq x\right \}}$