## Calculus 10th Edition

$c=3.$
$f(x)$ is continuous for all values of $x$ such that $x\ne1\to f(x)$ is continuous over the interval $(-\infty, 1)$ U $(1, \infty).$ $f(\frac{5}{2})=\dfrac{\frac{5}{2}(\frac{5}{2}+1)}{\frac{5}{2}-1}=\dfrac{35}{6}.$ $f(4)=\dfrac{4(4+1)}{4-1}=\dfrac{20}{3}.$ SInce $f(x)$ is continuous over the interval $[\frac{5}{2}, 4]$ and $\dfrac{35}{6}\lt6\lt\dfrac{20}{3},$ then the Intermediate Value Theorem guarantees the existence of at least one value $c$ such that $f(c)=6.$ $\dfrac{x^2+x}{x-1}=6\to x^2+x=6x-6\to x^2-5x+6=0\to$ $(x-2)(x-3)\to x=2$ or $x=3.$ $c=3$ (the other value is rejected since it is not in the specified domain.