Continuity

In calculus, a continuous function occurs when there or no gaps or breaks in the graph.

These breaks can take on several forms:

  • Holes
  • Linear function graph with a hole at x=1.
  • Jumps
  • Continuous function graph representing a jump between c=1 and c=3.
  • Vertical Asymptotes
  • Graph of 2 branches approching a vertical asymptote at x=1.

Additionally, 3 conditions need to be met for a function to be continuous:

  1. \(f(a)\) is defined
  2. \(\lim\limits_{x\to a}f(x)\) exists
  3. \(\lim\limits_{x\to a}f(x)=f(a)\)

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.

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:

\( \lim\limits_{x\to-1^-}-2x+a = \lim\limits_{x\to-1^+}x^2 + b\)

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:
$$f(x) = \left\{ \begin{array}{c} 2x+a,&x \le -1\\ 22,&-1 \lt x \le 3\\ -bx^2+31,&3 \lt x \le 5\\ \sqrt{cx}+b&5\lt x \end{array} \right. $$

Continuous vs Discontinuous

Try to determine which of the following expressions are continuous and discontinuous.

\(x^2\)

\(3x + 1\)

\(\cfrac{1}{x-1}\)

\(\cos(x)\)

\(\cfrac{6x^3}{x}\)

\(5\)

\(\tan(x)\)

\(8x^4\)

\(2^x + 5\)

\( \left\{ \begin{array}{c} x+2,&x \le 5\\ x^2-4,&5 \lt x \le 10\\ \end{array} \right. \)

Continuous

Disontinuous

Try these questions: