Answer
\[c=\frac{2}{3}\]
Work Step by Step
\[f(x)=\left\{\begin{array}{ll}
cx^2+2x\;\; , x<2\\
x^3-cx\;\;\; , x\geq 2
\end{array}\right.\]
Clearly $f(x)$ is continuous if $x\neq 2$. We will check the continuity at $r=2$.
\[\lim_{x\rightarrow 2^{-}}f(x)=\lim_{x\rightarrow 2}(cx^2+2x)=4c+4\]
\[\lim_{x\rightarrow 2^{+}}f(x)=\lim_{x\rightarrow 2}(x^3-cx)=8-2c\]
If $f$ is continuous at $x=2$ then:
\[\lim_{x\rightarrow 2^{-}}f(x)=\lim_{x\rightarrow 2^{+}}f(x)\]
\[\Rightarrow 4c+4=8-2c\Rightarrow 6c=4\Rightarrow c=\frac{2}{3}\]
Hence $f$ is continuous on $(-\infty,\infty)$ for $c=\displaystyle\frac{2}{3}$.
