Question

1. does the ivt apply to the function h(x) = \frac{x^{2}+x}{x - 2} on the interval 2.5, 5? if so, find the value of c guaranteed to exist, such that h(c) = 12.

Explanation:

Step1: Check function continuity

The function $h(x)=\frac{x^{2}+x}{x - 2}=\frac{x(x + 1)}{x - 2}$ is a rational - function. It is continuous on its domain. The domain of $h(x)$ is all real numbers except $x = 2$. Since the interval $[2.5,5]$ does not contain $x = 2$, $h(x)$ is continuous on $[2.5,5]$. So, the Intermediate - Value Theorem (IVT) applies.

Step2: Set up the equation

We want to find $c$ such that $h(c)=12$, so we set up the equation $\frac{c^{2}+c}{c - 2}=12$.
Cross - multiply to get $c^{2}+c=12(c - 2)$.
Expand the right - hand side: $c^{2}+c=12c-24$.
Rearrange to get a quadratic equation: $c^{2}+c-12c + 24 = 0$, which simplifies to $c^{2}-11c + 24 = 0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b=-11$, $c = 24$), we use the quadratic formula $c=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-11)^{2}-4\times1\times24=121 - 96 = 25$.
Then $c=\frac{11\pm\sqrt{25}}{2}=\frac{11\pm5}{2}$.
We have two solutions: $c_1=\frac{11 + 5}{2}=8$ and $c_2=\frac{11 - 5}{2}=3$.
Since $c$ must be in the interval $[2.5,5]$, we reject $c = 8$.

Answer:

$c = 3$