Find the values of k so that the function f is continuous at the indicated point.
$f(x)=\left\{\begin{array}{ll}k x^{2}, & \text { if } x \leq 2 \\ 3, & \text { if } x>2\end{array} \quad\right.$ at $x=2$
The given function is $f(x)= \begin{cases}k x^{2}, & \text { if } x \leq 2 \\ 3, & \text { if } x>2\end{cases}$
The given function f is continuous at x = 2, if f is defined at x = 2 and if the value of f at x = 2 equals the limit of f at x = 2
It is evident that $f$ is defined at $x=2$ and $f(2)=k(2)^{2}=4 k$
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2)$
$\Rightarrow \lim _{x \rightarrow 2^{-}}\left(k x^{2}\right)=\lim _{x \rightarrow 2^{+}}(3)=4 k$
$\Rightarrow k \times 2^{2}=3=4 k$
$\Rightarrow 4 k=3=4 k$
$\Rightarrow 4 k=3$
$\Rightarrow k=\frac{3}{4}$
Therefore, the required value of $k$ is $\frac{3}{4}$.
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