2.4 Continuity  (Page 6/16)

$\frac{2x^2-5x+3}{x-1}$ at $x=1$

$h\left(\theta \right)=\frac{\text{sin}\theta -\text{cos}\theta }{\text{tan}\theta }$ at $\theta =\pi$

$g\left(u\right)=\{\begin{array}{ll}\frac{6u^2+u-2}{2u-1}& \text{if}u\ne \frac{1}{2}\hfill \\ \frac{7}{2}\hfill & \text{if}u=\frac{1}{2}\hfill \end{array},$ at $u=\frac{1}{2}$

$f\left(y\right)=\frac{\text{sin}\left(\pi y\right)}{\text{tan}\left(\pi y\right)},$ at $y=1$

$f\left(x\right)=\{\begin{array}{ll}{x}^2-e^x& \text{if}x<0\hfill \\ x-1& \text{if}x\ge 0\hfill \end{array},$ at $x=0$

$f\left(x\right)=\{\begin{array}{l}x\text{sin}\left(x\right)\text{if}x\le \pi \\ x\text{tan}\left(x\right)\text{if}x>\pi \end{array},$ at $x=\pi$

$f\left(x\right)=\{\begin{array}{cc}3x+2,\hfill & x<k\hfill \\ 2x-3,\hfill & k\le x\le 8\hfill \end{array}$

$f\left(\theta \right)=\{\begin{array}{ll}\hfill \text{sin}\theta ,& 0\le \theta <\frac{\pi }{2}\hfill \\ \text{cos}\left(\theta +k\right),\hfill & \frac{\pi }{2}\le \theta \le \pi \hfill \end{array}$

$f\left(x\right)=\{\begin{array}{cc}\frac{x^2+3x+2}{x+2},\hfill & x\ne -2\hfill \\ \hfill k,& x=\mathrm{-2}\hfill \end{array}$

$f\left(x\right)=\{\begin{array}{cc}\hfill e^{kx},& 0\le x<4\hfill \\ x+3,\hfill & 4\le x\le 8\hfill \end{array}$

$f\left(x\right)=\{\begin{array}{cc}\hfill \sqrt{kx},& 0\le x\le 3\hfill \\ x+1,\hfill & 3<x\le 10\hfill \end{array}$

Let $h\left(x\right)=\{\begin{array}{ll}3x^2-4,& x\le 2\hfill \\ 5+4x,& x>2\hfill \end{array}$ Over the interval $\left[0,4\right],$ there is no value of x such that $h\left(x\right)=10,$ although $h\left(0\right)<10$ and $h\left(4\right)>10.$ Explain why this does not contradict the IVT.

A particle moving along a line has at each time t a position function $s\left(t\right),$ which is continuous. Assume $s\left(2\right)=5$ and $s\left(5\right)=2.$ Another particle moves such that its position is given by $h\left(t\right)=s\left(t\right)-t.$ Explain why there must be a value c for $2<c<5$ such that $h\left(c\right)=0.$

[T] Use the statement “The cosine of t is equal to t cubed.”

1. Write a mathematical equation of the statement.
2. Prove that the equation in part a. has at least one real solution.
3. Use a calculator to find an interval of length 0.01 that contains a solution.

Apply the IVT to determine whether $2^x=x^3$ has a solution in one of the intervals $\left[1.25,1.375\right]$ or $\left[1.375,1.5\right].$ Briefly explain your response for each interval.

Consider the graph of the function $y=f\left(x\right)$ shown in the following graph.

1. Find all values for which the function is discontinuous.
2. For each value in part a., state why the formal definition of continuity does not apply.
3. Classify each discontinuity as either jump, removable, or infinite.

Let $f\left(x\right)=\{\begin{array}{c}3x,x>1\\ x^3,x<1\end{array}.$

1. Sketch the graph of f .
2. Is it possible to find a value k such that $f\left(1\right)=k,$ which makes $f\left(x\right)$ continuous for all real numbers? Briefly explain.

Let $f\left(x\right)=\frac{x^4-1}{x^2-1}$ for $x\ne -1,1.$

1. Sketch the graph of f .
2. Is it possible to find values $k_1$ and $k_2$ such that $f\left(\mathrm{-1}\right)=k$ and $f\left(1\right)=k_2,$ and that makes $f\left(x\right)$ continuous for all real numbers? Briefly explain.

Sketch the graph of the function $y=f\left(x\right)$ with properties i. through vii.

1. The domain of f is $\left(\text{−}\infty ,\text{+}\infty \right).$
2. f has an infinite discontinuity at $x=\mathrm{-6}.$
3. $f\left(\mathrm{-6}\right)=3$
4. $\underset{x\to {\mathrm{-3}}^{-}}{\text{lim}}f\left(x\right)=\underset{x\to {\mathrm{-3}}^{+}}{\text{lim}}f\left(x\right)=2$
5. $f\left(\mathrm{-3}\right)=3$
6. f is left continuous but not right continuous at $x=3.$
7. $\underset{x\to -\infty }{\text{lim}}f\left(x\right)=\text{−}\infty$ and $\underset{x\to +\infty }{\text{lim}}f\left(x\right)=\text{+}\infty$

Sketch the graph of the function $y=f\left(x\right)$ with properties i. through iv.

1. The domain of f is $\left[0,5\right].$
2. $\underset{x\to 1^{+}}{\text{lim}}f\left(x\right)$ and $\underset{x\to 1^{-}}{\text{lim}}f\left(x\right)$ exist and are equal.
3. $f\left(x\right)$ is left continuous but not continuous at $x=2,$ and right continuous but not continuous at $x=3.$
4. $f\left(x\right)$ has a removable discontinuity at $x=1,$ a jump discontinuity at $x=2,$ and the following limits hold: $\underset{x\to 3^{-}}{\text{lim}}f\left(x\right)=\text{−}\infty$ and $\underset{x\to 3^{+}}{\text{lim}}f\left(x\right)=2.$