Question

1. find all values of (a,binmathbb{r}) so that the following function is continuous at (x = 2) and (x = 5). (g(x)=\begin{cases}a - e^{x - 2}&\text{if }x<2\sqrt{6 - x}&\text{if }2leq x<5\\-sqrt{b}cscleft(\frac{pi x}{4}

ight)&\text{if }xgeq5end{cases}) see examples 1.8.9 and 1.8.10 in chads calculus notes

Explanation:

Step1: Recall continuity condition

For a function to be continuous at a point $x = c$, $\lim_{x
ightarrow c^{-}}g(x)=\lim_{x
ightarrow c^{+}}g(x)=g(c)$. First, consider continuity at $x = 2$.
$\lim_{x
ightarrow 2^{-}}g(x)=\lim_{x
ightarrow 2^{-}}(a - e^{x - 2})=a - e^{2-2}=a - 1$.
$\lim_{x
ightarrow 2^{+}}g(x)=\lim_{x
ightarrow 2^{+}}\sqrt{6 - x}=\sqrt{6 - 2}=2$.
Since $g(x)$ is continuous at $x = 2$, we have $a-1 = 2$.

Step2: Solve for $a$

From $a - 1=2$, we add 1 to both sides of the equation: $a=2 + 1=3$.

Step3: Consider continuity at $x = 5$

Now, with $a = 3$, consider continuity at $x = 5$.
$\lim_{x
ightarrow 5^{-}}g(x)=\lim_{x
ightarrow 5^{-}}\sqrt{6 - x}=\sqrt{6 - 5}=1$.
$\lim_{x
ightarrow 5^{+}}g(x)=\lim_{x
ightarrow 5^{+}}(-\sqrt{b}\csc(\frac{\pi x}{4}))=-\sqrt{b}\csc(\frac{5\pi}{4})$.
Since $\csc(\frac{5\pi}{4})=\frac{1}{\sin(\frac{5\pi}{4})}=\frac{1}{-\frac{\sqrt{2}}{2}}=-\sqrt{2}$, then $\lim_{x
ightarrow 5^{+}}g(x)=-\sqrt{b}\times(-\sqrt{2})=\sqrt{2b}$.

Step4: Solve for $b$

Since $g(x)$ is continuous at $x = 5$, we set $\sqrt{2b}=1$. Square both sides: $2b = 1$, so $b=\frac{1}{2}$.

Answer:

$a = 3$, $b=\frac{1}{2}$