Calculus I - Continuity (Practice Problems)

Section 2.9 : Continuity

The graph of $f\left( x \right)$ is given below. Based on this graph determine where the function is discontinuous.
$The graph of $f\left( x \right)$ on $ - 6 \le x \le 5$. Starting at the left the graph increases to a peak at approximately $\left( { - 5,4} \right)$ and then decreases to a closed dot at $\left( { - 4,3} \right)$. There is an open dot at $\left( { - 4, - 2} \right)$ and from there the graph increases to a closed dot at $\left( { - 1,4} \right)$. The graph continues on from that point up to a peak at approximately $\left( {1,5} \right)$ and then falls steeply to a closed dot at $\left( {2, - 1} \right)$. There is an open dot at $\left( {2,5} \right)$ and from there the graph rises to a peak at approximately $\left( {3,6} \right)$ and then decreases to an open dot at $\left( {4,2} \right)$. The graph then continues on down from that point, leaving the bottom of the graph at approximately $x = 5$.$
Solution
The graph of $f\left( x \right)$ is given below. Based on this graph determine where the function is discontinuous.
$The graph of $f\left( x \right)$ on $ - 12 \le x \le 12$. Coming down from the upper left the graph decreases to an open dot at the bottom of a valley at approximately $\left( { - 8, - 6} \right)$, with a separate closed dot sitting above it at approximately $\left( { - 8, - 3} \right)$. From the open dot the graph increases, crossing the x-axis near $x = - 4$, and rises to a closed dot at approximately $\left( { - 2,3} \right)$ with the graph continuing up off the top of the picture just to the right of that point. The graph picks back up near $\left( {0,1} \right)$, rises to a small peak at approximately $\left( {3,3} \right)$, dips slightly and then runs out to an open dot at $\left( {6,2} \right)$, with a separate closed dot sitting above it at $\left( {6,5} \right)$. From the closed dot the graph decreases through a closed dot at approximately $\left( {10,1} \right)$ and continues on down to the lower right.$
Solution

For problems 3 – 7 using only Properties 1 – 9 from the Limit Properties section, one-sided limit properties (if needed) and the definition of continuity determine if the given function is continuous or discontinuous at the indicated points.

$\displaystyle f\left( x \right) = \frac{{4x + 5}}{{9 - 3x}}$
1. $x = - 1$
2. $x = 0$
3. $x = 3$
Solution
$\displaystyle g\left( z \right) = \frac{6}{{{z^2} - 3z - 10}}$
1. $z = - 2$
2. $z = 0$
3. $z = 5$
Solution
$g\left( x \right) = \left\{ {\begin{array}{rl}{2x}&{x < 6}\\{x - 1}&{x \ge 6}\end{array}} \right.$
1. $x = 4$
2. $x = 6$
Solution
$h\left( t \right) = \left\{ {\begin{array}{rl}{{t^2}}&{t < - 2}\\{t + 6}&{t \ge - 2}\end{array}} \right.$
1. $t = - 2$
2. $t = 10$
Solution
$g\left( x \right) = \left\{ {\begin{array}{rc}{1 - 3x}&{x < - 6}\\7&{x = - 6}\\{{x^3}}&{ - 6 < x < 1}\\1&{x = 1}\\{2 - x}&{x > 1}\end{array}} \right.$
1. $x = - 6$
2. $x = 1$
Solution

For problems 8 – 12 determine where the given function is discontinuous.

$\displaystyle f\left( x \right) = \frac{{{x^2} - 9}}{{3{x^2} + 2x - 8}}$ Solution
$\displaystyle R\left( t \right) = \frac{{8t}}{{{t^2} - 9t - 1}}$ Solution
$\displaystyle h\left( z \right) = \frac{1}{{2 - 4\cos \left( {3z} \right)}}$ Solution
$\displaystyle y\left( x \right) = \frac{x}{{7 - {{\bf{e}}^{2x + 3}}}}$ Solution
$g\left( x \right) = \tan \left( {2x} \right)$ Solution

For problems 13 – 15 use the Intermediate Value Theorem to show that the given equation has at least one solution in the indicated interval. Note that you are NOT asked to find the solution only show that at least one must exist in the indicated interval.

$25 - 8{x^2} - {x^3} = 0$ on $\left[ { - 2,4} \right]$ Solution
${w^2} - 4\ln \left( {5w + 2} \right) = 0$ on $\left[ {0,4} \right]$ Solution
$4t + 10{{\bf{e}}^t} - {{\bf{e}}^{2t}} = 0$ on $\left[ {1,3} \right]$ Solution