# How do you verify the intermediate value theorem over the interval [5/2,4], and find the c that is guaranteed by the theorem such that f(c)=6 where f(x)=(x^2+x)/(x-1)?

Oct 20, 2016

Please see the verification in the explanation section.

#### Explanation:

$f$ is continuous on $\left[\frac{5}{2} , 4\right]$

Proof : $f$ is a rational function and rational functions are continuous on their domains. $f$ is continuous at all reals except $1$. (Its domain is all reals except $1$.) Since $1$ is not is $\left[\frac{5}{2} , 4\right]$, $f$ is continuous on that closed interval.

$6$ is between $f \left(\frac{5}{2}\right)$ and #f(4)

Proof: $f \left(\frac{5}{2}\right) = \frac{35}{6} < \frac{36}{6} = 6$ and $f \left(4\right) = \frac{20}{3} > \frac{18}{3} = 6$

Therefore, there is a $c$ in $\left(\frac{5}{2} , 4\right)$ with $f \left(c\right) = 6$

To find the $c$ (or $c$'s), solve the equation:

$f \left(x\right) = 6$.

Discard values outside $\left(\frac{5}{2} , 4\right)$.

Note the solutions are $2$ and $3$, but only $3$ is in the interval.