Consider the function \(h(x)=\begin{cases}\dfrac{x^2-9}{x-3}, & \text {for } x \neq 3 \\ k, & \text {for } x=3\end{cases}\)
- For what value of \(k\) is \(h(x)\) continuous at \(x =3\)? (2 marks)
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- Sketch \(y=h (x)\) for this value of \(k\). (2 marks)
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a. \(\text{Since} \ \ \dfrac{x^2-9}{x-3}=\dfrac{(x+3)(x-3)}{x-3}=x+3\)
\(\text{As} \ \ x \rightarrow 3, x+3 \rightarrow 6\)
\(h(x) \ \text {is continuous when}\ \ k=6\)
b.
Show Worked Solution
a. \(\text{Since} \ \ \dfrac{x^2-9}{x-3}=\dfrac{(x+3)(x-3)}{x-3}=x+3\)
\(\text{As} \ \ x \rightarrow 3, x+3 \rightarrow 6\)
\(h(x) \ \text {is continuous when}\ \ k=6\)
b.