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Solve the following.
Determine the value of k for which the following function is continuous at x = 3. f(x) = {((x + 3)²…
Determine the value of $k$ for which the following function is continuous at $x = 3$.
$f(x) = \begin{cases} \frac{(x + 3)^2 - 36}{x - 3} & ; x \neq 3 \\ k & ; x = 3 \end{cases}$
Answer
Answer
For continuity at $x = 3$: $\lim_{x\to3} f(x) = f(3) = k$.
$\frac{(x+3)^2 - 36}{x-3} = \frac{(x+3-6)(x+3+6)}{x-3} = \frac{(x-3)(x+9)}{x-3} = x + 9$ for $x \neq 3$.
$\lim_{x\to3}(x+9) = 12$
Hence $k = 12$.
Final answer: 12