## Calculus: Early Transcendentals 8th Edition

$c = \frac{2}{3}$
Clearly the graph is continuous on the intervals $(-\infty,2)$ and $[2,\infty)$ We need to find the value $c$ such that the graph is continuous at the point $x=2$.: $cx^2+2x = x^3-cx$ $c(2)^2+2(2) = (2)^3-c(2)$ $4c+4 = 8-2c$ $6c = 4$ $c = \frac{2}{3}$