## Calculus with Applications (10th Edition)

$$f(x)=\left\{\begin{array}{ll}{k x^{2}} & {\text { if } x \leq 2} \\ {x+k} & {\text { if } x>2}\end{array}\right.$$ The value of the constant $k$ that makes the function continuous is . $k=\frac{2}{3}$
$$f(x)=\left\{\begin{array}{ll}{k x^{2}} & {\text { if } x \leq 2} \\ {x+k} & {\text { if } x>2}\end{array}\right.$$ Since each piece of this function is a polynomial, the only $x$-value where $f$ might be discontinuous here is 2. We investigate at $x=2$ first. From the left, where $x$-values are less than $2$, $$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}\left(kx^{2}\right)=k(2)^{2}=4k$$ From the right, where $x$-values are greater than 2, $$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}\left(x+k\right)=2+k$$ Furthermore, $$f(2)=k (2)^{2}=4k$$ since, $f$ is continuous, then we have $$f(2)=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{-}}f(x)$$ so that $$k x^{2}=x+k \text { for } x=2$$ we obtain \begin{aligned} k(2)^{2} &=2+k \\ 4 k &=2+k \\ 3 k &=2 \\ k &=\frac{2}{3} . \end{aligned} Thus the value of the constant $k$ that makes the function continuous is . $k=\frac{2}{3}$