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: Identifying Points of Discontinuity
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
- a) For \( x = 0 \), the denominator of function \( f(x) \) is equal to \( 0 \), so \( f(x) \) is not defined and does not have a limit. Therefore, \( f(x) \) is discontinuous at \( x = 0 \).
- b) For \( x = 2 \), the denominator of function \( g(x) \) is equal to \( 0 \) and is undefined with no limit. Function \( g(x) \) is not continuous at \( x = 2 \).
- c) The denominator of \( h(x) \) factors as \( x^2 - 1 = (x - 1)(x + 1) \). The denominator is zero for \( x = 1 \) and \( x = -1 \), where the function is undefined and has no limits. Function \( h \) is discontinuous at \( x = 1 \) and \( x = -1 \).
- d) \( \tan(x) \) is undefined for all values of \( x \) such that \( x = \frac{\pi}{2} + k\pi \) (where \( k \) is any integer), making it discontinuous at these values.
- e) The denominator of \( j(x) \) equals zero when \( \cos(x) - 1 = 0 \), or \( x = 2k\pi \) where \( k \) is an integer. The function is undefined and discontinuous at these points.
- f) Function \( k(x) \) is the ratio of two continuous functions with a denominator \( x^2 + 5 \) that is never equal to \( 0 \). It is defined for all real values of \( x \) and has no points of discontinuity.
- g) \( l(x) = \dfrac{x + 4}{x + 4} = 1 \) for \( x \ne -4 \). Since \( \lim_{x \to -4} l(x) = 1 = l(-4) \), function \( l(x) \) is continuous for all real values of \( x \) and has no point of discontinuity.
Example 2: Finding Parameters for Piecewise Continuity
Find \( b \) such that \( f(x) \) given below is continuous:
\[ f(x) = \begin{cases} 2x^2+b & x \ge -1 \\ -x^3 & x < -1 \end{cases} \]Solution to Example 2
For \( x > -1 \), \( f(x) = 2x^2 + b \) is a polynomial function and therefore continuous.
For \( x < -1 \), \( f(x) = -x^3 \) is a polynomial function and therefore continuous.
At \( x = -1 \):
\[ f(-1) = 2(-1)^2 + b = 2 + b \]Consider the left-hand and right-hand limits:
\[ L_1 = \lim_{x\to -1^{\text{--}}} f(x) = -(-1)^3 = 1 \] \[ L_2 = \lim_{x\to -1^+} f(x) = 2(-1)^2 + b = 2 + b \]For function \( f \) to be continuous at \( x = -1 \), we require \( L_2 = L_1 \):
\[ 2 + b = 1 \implies b = -1 \]Substituting \( b = -1 \) into the function gives:
\[ f(x) = \begin{cases} 2x^2-1 & x \ge -1 \\ -x^3 & x < -1 \end{cases} \]The graph of \( f \) is shown below, confirming continuity at \( x = -1 \):
Example 3: Continuity of Function and its Derivative
Find \( a \) and \( b \) such that both \( g(x) \) given below and its first derivative are continuous:
\[ g(x) = \begin{cases} ax^2+b & x \ge 2 \\ -2x+2 & x < 2 \end{cases} \]Solution to Example 3
Continuity of function \( g \):
- For \( x > 2 \), \( g(x) = ax^2 + b \) is continuous.
- For \( x < 2 \), \( g(x) = -2x + 2 \) is continuous.
Limits at \( x = 2 \):
\[ L_1 = \lim_{x\to 2^+} g(x) = a(2)^2 + b = 4a + b \] \[ L_2 = \lim_{x\to 2^{\text{--}}} g(x) = -2(2) + 2 = -2 \]For continuity at \( x = 2 \), we must have \( L_1 = L_2 \):
\[ 4a + b = -2 \]Continuity of the derivative \( g' \):
- For \( x > 2 \), \( g'(x) = 2ax \) is continuous.
- For \( x < 2 \), \( g'(x) = -2 \) is continuous.
Derivative limits at \( x = 2 \):
\[ l_1 = \lim_{x\to 2^+} g'(x) = 2a(2) = 4a \] \[ l_2 = \lim_{x\to 2^{\text{--}}} g'(x) = -2 \]For continuity of \( g' \) at \( x = 2 \), we must have \( l_1 = l_2 \):
\[ 4a = -2 \implies a = -\frac{1}{2} \]Substitute \( a = -\frac{1}{2} \) into \( 4a + b = -2 \):
\[ 4\left(-\frac{1}{2}\right) + b = -2 \implies -2 + b = -2 \implies b = 0 \]Thus, the function is:
\[ g(x) = \begin{cases} -\dfrac{1}{2}x^2 & x \ge 2 \\ -2x+2 & x < 2 \end{cases} \]Function \( g(x) \) is graphed below, showing that both the function and its derivative (slope) are continuous at \( x = 2 \):
