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Consider the function
\begin{align*}
f(x)=\begin{cases}2^x, & \text {for }\ x<0 \\ m x+c, & \text {for }\ 0 \leq x \leq 2 \\ \dfrac{8}{x}, & \text {for }\ x>2\end{cases}
\end{align*}
Given that \(f(x)\) is continuous at both \(x =0\) and \(x =2\):
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a. \(f(x)=\left\{\begin{array}{cl}2^x & \text {for } x<0 \\ m x+c & \text {for } 0 \leq x \leq 2 \\ \dfrac{8}{x} & \text {for } x>2\end{array}\right.\)
\(\text {Continuous at}\ \ x=0:\)
\(2^\circ=m(0)+c \ \Rightarrow \ c=1\)
\(\text {Continuous at}\ \ x=2:\)
\(2 m+1=\dfrac{8}{2} \ \Rightarrow \ m=\dfrac{3}{2}\)
b. \(\text{Asymptotes:}\)
\(\text{As} \ x \rightarrow-\infty, 2^x \rightarrow 0^{+}\)
\(\text{As} \ x \rightarrow \infty, \dfrac{8}{x} \rightarrow 0^{+}\)
\(\text{Asymptote at} \ \ y=0.\)
\(\text{There are no vertical asymptotes.}\)
a. \(f(x)=\left\{\begin{array}{cl}2^x & \text {for } x<0 \\ m x+c & \text {for } 0 \leq x \leq 2 \\ \frac{8}{x} & \text {for } x>2\end{array}\right.\)
\(\text {Continuous at}\ \ x=0:\)
\(2^\circ=m(0)+c \ \Rightarrow \ c=1\)
\(\text {Continuous at}\ \ x=2:\)
\(2 m+1=d\dfrac{8}{2} \ \Rightarrow \ m=\dfrac{3}{2}\)
b. \(\text{Asymptotes:}\)
\(\text{As} \ x \rightarrow-\infty, 2^x \rightarrow 0^{+}\)
\(\text{As} \ x \rightarrow \infty, \dfrac{8}{x} \rightarrow 0^{+}\)
\(\text{Asymptote at} \ \ y=0.\)
\(\text{There are no vertical asymptotes.}\)