Question

1. find the value(s) of the constant k that would make the following function continuous at x = 4. show all work!!

g(x)=\begin{cases}e^{kx},&\text{if }0leq x<4\\x + 3,&\text{if }4leq xleq8end{cases}
lim_{x
ightarrow4^{-}}e^{kx}=lim_{x
ightarrow4^{+}}x + 3
e^{k(4)}=4 + 3

Explanation:

Step1: Recall continuity condition

For a function to be continuous at $x = a$, $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)$. Here $a = 4$, so $\lim_{x
ightarrow4^{-}}g(x)=\lim_{x
ightarrow4^{+}}g(x)$. Given $g(x)=

$$\begin{cases}e^{kx},&0\leq x < 4\\x + 3,&4\leq x\leq8\end{cases}$$

$, we have $\lim_{x
ightarrow4^{-}}e^{kx}=\lim_{x
ightarrow4^{+}}(x + 3)$.

Step2: Evaluate the limits

Substitute $x = 4$ into the expressions for the limits. $\lim_{x
ightarrow4^{-}}e^{kx}=e^{4k}$ and $\lim_{x
ightarrow4^{+}}(x + 3)=4 + 3=7$. So we get the equation $e^{4k}=7$.

Step3: Solve for $k$

Take the natural - logarithm of both sides of the equation $e^{4k}=7$. Using the property $\ln(e^{u})=u$, we have $\ln(e^{4k})=\ln(7)$. So $4k=\ln(7)$. Then $k=\frac{\ln(7)}{4}$.

Answer:

$k=\frac{\ln(7)}{4}$