# How do you determine all values of c that satisfy the mean value theorem on the interval [2,5] for f(x) = 1 / (x-1)?

Apr 4, 2016

$\forall c \setminus \in \left[f \left(5\right) , f \left(2\right)\right] = \left[\frac{1}{4} , 1\right]$, $\exists y \setminus \in \left[2 , 5\right]$ such that $f \left(y\right) = c$.
Explanation below.

#### Explanation:

Since $f$ is a quotient of continuous function :
$f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)} , g \left(x\right) = 1$ and $h \left(x\right) = x - 1$,
it is defined and continuous for all $x \setminus \in \mathbb{R} \setminus b a c k s l a s h \left\{1\right\}$ ($f \left(1\right)$ is not defined because we would divide by zero).

Therefore, the mean value theorem is satisfied for all closed interval $\left[a , b\right] \setminus \subset \mathbb{R} \setminus b a c k s l a s h \left\{1\right\}$.

Since $\left[2 , 5\right] \setminus \subset \mathbb{R} \setminus b a c k s l a s h \left\{1\right\}$ and $\left[2 , 5\right]$ is closed, we have, by the mean value theorem, that $\forall c \setminus \in \left[f \left(5\right) , f \left(2\right)\right] = \left[\frac{1}{4} , 1\right]$, $\exists y \setminus \in \left[2 , 5\right]$ such that $f \left(y\right) = c$.

Apr 4, 2016

$c = 3 \mathmr{and} c = - 1$

#### Explanation:

$f ' \left(c\right) = \frac{f \left(b\right) - f \left(a\right)}{b - a}$

$f ' \left(c\right) = \frac{f \left(5\right) - f \left(2\right)}{5 - 2}$

$- \frac{1}{c - 1} ^ 2 = \frac{\frac{1}{4} - 1}{3} = \frac{- \frac{3}{4}}{3} = - \frac{3}{4} \cdot \frac{1}{3} = - \frac{1}{4}$

$- 1 = - \frac{1}{4} {\left(c - 1\right)}^{2}$

$4 = {c}^{2} - 2 c + 1$

$0 = {c}^{2} - 2 c - 3$

$\left(c - 3\right) \left(c + 1\right) = 0$

$c = 3 \mathmr{and} c = - 1$