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Question
Determine the value of the constant k so that the function
\[f\left( x \right) = \begin{cases}k x^2 , if & x \leq 2 \\ 3 , if & x > 2\end{cases}\text{is continuous at x} = 2 .\]
Solution
\[f\left( x \right) = \binom{k x^2 , if x \leq 2}{3, if x > 2}\]
If
\[f\left( x \right)\] is continuous at x = 2, then
\[\lim_{x \to 2^-} f\left( x \right) = \lim_{x \to 2^+} f\left( x \right) = f\left( 2 \right)\]
Now
\[\lim_{x \to 2^-} f\left( x \right) = \lim_{h \to 0} f\left( 2 - h \right) = \lim_{h \to 0} k \left( 2 - h \right)^2 = 4k\]
\[f\left( 2 \right) = 3\]
From (1), we have
\[4k = 3\]
\[ \Rightarrow k = \frac{3}{4}\]
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