In calculus, a continuous function occurs when there or no gaps or breaks in the graph.
These breaks can take on several forms:
Holes
Jumps
Vertical Asymptotes
Additionally, 3 conditions need to be met for a function to be continuous:
\(f(a)\) is defined
\(\lim\limits_{x\to a}f(x)\) exists
\(\lim\limits_{x\to a}f(x)=f(a)\)
We can instantly tell if there is a discontinuity if there is an undefined value. This can be indicated either by \(\sqrt{\text{neg}}\) or \(\cfrac{\text{number}}{0}\).
We can determine whether or not a function is continuous either algebraically or graphically We can do so algebraically by checking the function at any breaks. We can do so graphically by inspecting the graph for any breaks.
Example
Determine if the function, \(f(x) = x^3 - x\), is continuous. If not, identify where the discontinuity is.
This function is continuous since there aren't any square roots or denominators to make it undefined.
Determine if the function, \(h(x) = \cfrac{2x+2}{3x-12}\) is continuous. If not, identify where the discontinuity is. Also, show a graphical representation.
Show Answer
We can determine that this function is dicontinuous since there is an \(x\)-value in the denominator.
In order to determine where it's dicontinuous, we can factor out the common factors from both the numerator and denominator:
\(h(x) = \cfrac{2x+2}{3x-12}\)
\(h(x) = \cfrac{2(x+1)}{3(x-4)}\)
From this factored function, we can determine the Vertical Asymptote as \(4\). As a result, the function is discontinuous at these respective values.
Before graphing this expression, we can determine the x and y-intercepts. In order to find the x-intercept, we can set the numerator equal to \(0\) and solve for \(x\):
\(\text{x-int}: x = -1\)
In order to find the \(y\)-intercept, we can substitute \(x\) with \(0\):
\(\text{x-int}: y = -\cfrac{1}{6}\)
Therefore, we can determine the \(x\) and \(y\)-intercepts as \(-1\) and \(-\cfrac{1}{6}\) respectively.
Now that we have determined all the pertinent values, we can graph this function to verify that it's not continuous. Since there is a Vertical Asymptote we can determine that this function is NOT continuous:
Example
Find the values of each constant that would make this function continuous:
$$f(x) =
\left\{
\begin{array}{c}
-2x+a,&x \le -1\\
x^2+b,&-1 \lt x \le 2\\
\cfrac{1}{x}+2a,&2 \lt x
\end{array}
\right.
$$
In order to verify the continuity of this function, we need to check each piece! A line and parabola are continuous. Thus, we only have to check at the value where the graph switches to see if there is a break.
The graph changes from the line to the parabola at \(x=-1\). So, we want the function to graphs to approach each other at that point:
As a result, we can substitute \(-1\) as \(x\) for both equations to get a new expression:
\(-2x + a = x^2 + b\)
\(-2(-1) + a = (-1)^2 + b\)
\(2 + a = 1 + b\)
\(b = a + 2 - 1\)
\(b = a + 1\)
Next, we check the third function. It is rational and has a VA at \(x = 0\) which would make it discontinuous. However, this graph is only valid on the domain \(x > 2\). On this domain, it is continuous.
Let's check the point where the parabola switches to the rational function at \(x=2\):
\(\lim\limits_{x\to2^-}x^2 + b = \lim\limits_{x\to2^+}\cfrac{1}{x}+2a\).
As a result, we can substitute \(2\) for \(x\) with both equations to get a new expression. We can also substitute \(a + 1\) for \(b\):
\(x^2+b = \cfrac{1}{x}+2a\)
\((2)^2+(a + 1) = \cfrac{1}{2}+2a\)
\(4+a+1 = \cfrac{1}{2}+2a\)
\(2a-a = 4 + 1 -\cfrac{1}{2}\)
\(\textcolor{red}{a = 4.5}\)
We can now determine \(b\) by subtituting \(4.5\) for \(a\) in \(b = a + 1\):
\(b = a + 1\)
\(b = 4.5 + 1\)
\(\textcolor{green}{b = 5.5}\)
Therefore, we can determine that \(\textcolor{red}{a = 4.5}\) and \(\textcolor{green}{b = 5.5}\). These values will make the function continuous.
Find the values of each constant that would make this function continuous:
$$f(x) =
\left\{
\begin{array}{c}
2x+a,&x \le -1\\
22,&-1 \lt x \le 3\\
-bx^2+31,&3 \lt x \le 5\\
\sqrt{cx}+b&5\lt x
\end{array}
\right.
$$
Show Answer
In order to verify the continuity of this function, we need to check every piece!
We will need to check each piece on its respective domain and where it switches functions. We can do this by setting the pieces equal to each other based on their domains.
From this example we can first determine that: \( \lim\limits_{x\to-1^-}2x+a = \lim\limits_{x\to-1^+}22\).
As a result, we can set both equations equal to each other and substitute \(-1\) for \(x\):
Therefore, we can determine that \(\textcolor{red}{a = 24}\), \(\textcolor{green}{b = 1}\), and \(\textcolor{blue}{c = 5}\). These values will make the function continuous.
Continuous vs Discontinuous
Try to determine which of the following expressions are continuous and discontinuous.