In the following exercise, find the value(s) of k that makes the function continuous over the given interval.

$$f(x)=\left\{\begin{array}{cl} e^{k x}, & 0 \leq x<4 \\ x+3, & 4 \leq x \leq 8 \end{array}\right.$$

For following Exercise find the value of the constant $k$ that makes the function continuous.

$$h(x)=\left\{\begin{array}{ll}{\frac{3 x^{2}+2 x-8}{x+2}} & {\text { if } x \neq-2} \\ {3 x+k} & {\text { if } x=-2}\end{array}\right.$$

The derivative of 2x - x^(-2) is

(A) 2 + 2x

(B) 2 - x^(-1)

(C) x^2 + 2x^(-3)

(D) 2 + 2x^(-3)

(E) 2 + 2x^(-1)

Find the value of $k$ that makes $16 x^2-k x+9=0$ a perfect square trinomial.

Continuous extension Find a value of $c$ that makes the function

$$f(x)= \begin{cases}\frac{9 x-3 \sin 3 x}{5 x^3}, & x \neq 0 \\ c, & x=0\end{cases}$$

continuous at $x=0$. Explain why your value of $c$ works.

In the following exercise, find the value of the constant k that makes the function continuous.

$$g(x)= \begin{cases}\frac{2 x^2-x-15}{x-3} & \text { if } x \neq 3 \\ k x-1 & \text { if } x=3\end{cases}$$

Suppose pure-wavelength light falls on a diffraction grating. What happens to the interference pattern if the same light falls on a grating that has more lines per centimeter? What happens to the interference pattern if a longer-wavelength light falls on the same grating? Explain how these two effects are consistent in terms of the relationship of wavelength to the distance between slits.

In the following exercise, find the value of the constant k that makes the function continuous.

$$g(x)= \begin{cases}\frac{\left(2 x^2-3 x-9\right)}{(x-3)} & \text { if } x \neq 3 \\ k x-12 & \text { if } x=3\end{cases}$$

For following Exercise find the value of the constant $k$ that makes the function continuous.

$$g(x)=\left\{\begin{array}{ll}{\frac{2 x^{2}-x-15}{x-3}} & {\text { if } x \neq 3} \\ {k x-1} & {\text { if } x=3}\end{array}\right.$$

In the following exercise, find the value(s) of k that makes the function continuous over the given interval.

$$f(x)=\left\{\begin{aligned} \sqrt{k x}, & 0 \leq x \leq 3 \\ x+1, & 3<x \leq 10 \end{aligned}\right.$$

Find the value of k that makes the following function continuous. Graph the function.

$$f(x)=\left\{\begin{array}{l}x^{2}-k, \text { if } x<-1 \\ 2 x-1, \text { if } x \geq-1\end{array}\right.$$

Find the value of $k$ such that

$$f(x)= \begin{cases}k x^2-k, & x \geq 3 \\ 4, & x<3 \text { is continuous for all real }\end{cases}$$

values of $x$.

Find the value of the constant $k$ that makes the function continuous:

$$\begin{align*} g(x)=\begin{cases} \dfrac{2x^2-5x-25}{x-5},&\text{ if }x\not=5\\ kx-5,&\text{ if }x=5. \end{cases} \end{align*}$$

a. Write the definitions of continuity at a point and continuity on a closed interval.

b. For the piecewise function in Figure 2$7 \mathrm{~d}$ make a table showing these quantities for $1,2,3,4$, and 5 , or stating that the quantity does not exist.

$$\begin{array}{ll}-f(c) & -\lim _{x \rightarrow c} f(x) \\ \text { - } \lim _{x \rightarrow c^{+}} f(x) & -\lim _{x \rightarrow c} f(x)\end{array}$$

• Continuity or kind of discontinuity

$$f(x)= \begin{cases}2+\frac{2}{|x-1|}, & \text { if } x<3 \text { and } x \neq 2 \\ 1, & \text { if } x=2 \\ 3+2 \cos \frac{\pi}{2}(x-3), & \text { if } 3 \leq x<5 \\ 4-3(x-6)^2, & \text { if } x \geq 5\end{cases}$$

c. Sketch graphs of the functions described.

• Has a removable discontinuity at $x=1$
• Has a step discontinuity at $x=2$
• Has a vertical asymptote at $x=3$
• Has a cusp at $x=4$
• Is continuous at $x=5$
• Has a limit as $x \rightarrow 6$ and a value of $f(6)$, but is discontinuous at $x=6$
• Has a left limit of $-2$ and a right limit of 5 as $x \rightarrow 1$

d. For the piecewise function

$$f(x)= \begin{cases}x^2, & \text { if } x \leq 2 \\ x^2-6 x+k_3 & \text { if } x>2\end{cases}$$

• Sketch the graph of $f$ if $k=10$.
• Show that $f$ is discontinuous at $x=2$ if $k=10$.
• Find the value of $k$ that makes $f$ continuous at $x=2$.