Continuity of piecewise functions

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Here we use limits to ensure piecewise functions are continuous.

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

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

Constant function
Polynomials
Rational functions
Power function
Exponential function
Logarithmic function
Trig functions
Inverse trig functions

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

We proved continuity of polynomials earlier using the Sum Law, Product Law and continuity of power functions.
We proved continuity of rational functions earlier using the Quotient Law and continuity of polynomials.
We can prove continuity of the remaining four trig functions using the Quotient Law and continuity of sine and cosine functions.

Since a continuous function and its inverse have ”unbroken” graphs, it follows that an inverse of a continuous function is continuous on its domain.
This implies that inverse trig functions are continuous on their domains.

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 $x<0$ and $x\ne 1$,}\\ \frac {x+1}{x^2+1} + c &\text {if $x\ge 0$}. \end {cases}$ Find $c$ 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 $\lim _{x\to 0^-} f(x) = \lim _{x\to 0^+}f(x) = f(\answer [given]{0}).$ In this case $\begin{align*} \lim_{x\to 0^-} f(x) &= \lim_{x\to 0^-}\answer[given]{\frac{x}{x-1}}\\ &= \answer[given]{\frac{0}{-1}}\\ &=\answer[given]{0}. \end{align*}$

On there other hand

$\begin{align*} \lim_{x\to 0^+} f(x) &= \lim_{x\to 0^+}\left(\answer[given]{\frac{x+1}{x^2+1} + c}\right)\\ &= \answer[given]{\frac{0+1}{0^2+1} + c}\\ &= \answer[given]{1+ c} \end{align*}$

Hence for our function to be continuous, we need $1 + c = 0\qquad \text {so}\qquad c = \answer [given]{-1}.$ Now, $\lim _{x\to 0} f(x) = f(\answer [given]{0})$ , and so $f$ is continuous.

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 $a$ 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 $f$ is continuous at $x=1$ , we need to find $a$ and $b$ such that $\lim _{x\to 1^-} f(x) = \lim _{x\to 1^+}f(x)=f(\answer [given]{1}).$ 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

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

Hence for this function to be continuous at $x=1$ , we must have that

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

Hmmmm. More work needs to be done.

To find $a$ and $b$ that make $f$ is continuous at $x=3$ , we need to find $a$ and $b$ such that $\lim _{x\to 3^-} f(x) =\lim _{x\to 3^+}f(x) =f(\answer [given]{3}).$ Looking at the limit from the left, we have

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

Looking at the limit from the right, we have

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

Hence for this function to be continuous at $x=3$ , we must have that

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

So now we have two equations and two unknowns: $3=a+b \qquad \text {and}\qquad a\cdot 2 + b -4 = 0.$ Set $b = 3-a$ and write

$\begin{align*} 0&= a\cdot 2 + (3-a) -4 \\ &= a -1, \end{align*}$

hence $a=\answer [given]{1}\qquad \text {and so}\qquad b =\answer [given]{2}.$ Let’s check, so now plugging in 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 $y=f(x)$ : $\graph [xmin=-10,xmax=10,ymin=0,ymax=30]{y=x+4\left \{x<1\right \},y=x^2+2x+2\left \{1\leq x<3\right \},y=6x-1\left \{3 \leq x\right \}}$