Questions on Continuity with Solutions
Questions with answers on the continuity of functions with emphasis on rational and piecewise functions. The continuity of a function and its derivative at a given point is discussed. Graphical meaning and interpretation of continuity are also included.
Example 1: For what values of x are each of the following functions discontinuous?
a) \;\; f(x) = \dfrac{1}{x} \;\; b) \;\; g(x) = \dfrac{2}{x-2} \;\; c) \;\; h(x) = \dfrac{x+1}{x^2-1}
d) \;\; i(x) = \tan(x) \;\; e) \;\; j(x) = \dfrac{1}{\cos(x)-1} \;\; f) \;\; k(x) = \dfrac{x+2}{x^2+5}
g) \;\; l(x) = \begin{cases}
\dfrac{x+4}{x+4} & x \ne -4 \\
1 & x = -4 \\
\end{cases}
Solution to Example 1
Example 2: Find b such that f(x) given below is continuous?
f(x) = \begin{cases}
2x^2+b & x \ge -1 \\
-x^3 & x \textless -1 \\
\end{cases}
Solution to Example 2
L1 = \lim_{x\to\ -1^-} f(x) = -(-1)^3 = 1
limit from right of -1
L2 = \lim_{x\to\ -1^+} f(x) = 2(-1)^2 + b = 2 + b
For function f to be continuous, we need to have L1 = L2 = 2 + b or 2 + b = 1 or b = -1. Substitute b by -1 in the given function to obtain
f(x) = \begin{cases}
2x^2-1 & x \ge -1 \\
-x^3 & x \textless -1 \\
\end{cases}
The graph of f is shown below and it is clear that the function is continuous at x = -1.
![]() Example 3: Find a and b such that both g(x) given below and its first derivative are continuous?
g(x) = \begin{cases}
ax^2+b & x \ge 2 \\
-2x+2 & x \textless 2 \\
\end{cases}
Solution to Example 3
L1 = \lim_{x\to\ 2^+} g(x) = a (2)^2 + b = 4 a + b
L2 = \lim_{x\to\ 2^-} g(x) = -2(2) + 2 = -2
For continuity of g at x = 2, we need to have L1 = L2 = g(2) Which gives 4 a + b = -2 Continuity of the derivative g' For x > 2, g '(x) = 2 a x is a polynomial function and therefore continuous. For x < 2, g '(x) = -2 is a constant function and therefore continuous. Let
l1 = \lim_{x\to\ 2^+} g'(x) = 2a(2) = 4 a
l2 = \lim_{x\to\ 2^-} g'(x) = -2
For continuity of g' at x = 2, we need to have l1 = l2 or 4 a = - 2 The last equation gives: a = - 1 / 2. And substitute a by - 1 / 2 in the equation 4 a + b = -2 obtained above, we obtain b = 0. Substitute a and b by their values to obtain:
g(x) = \begin{cases}
-\dfrac{1}{2}x^2 & x \ge 2 \\
-2x+2 & x \textless 2 \\
\end{cases}
Function g(x) is graphed below and it is clear that both the function and its derivative (slope) are continuous at x = 2.
![]() More on Continuous Functions in Calculus and Continuity Theorems and Their use in Calculus |