Finding $c$
Rolle's theorem and the Mean Value Theorem say that $f'(x)$ takes
on a certain value at some point $c$ between $a$ and $b$. But how
do you find such a $c$? Is there only one?
Example 1: Suppose that $f(x) = x^3 6x^2 + 5x$. Plugging
in, we see that $f(0)=0$ and $f(1)=0$. (The third root is $x=5$,
by the way). By Rolle's theorem, there has to be a $c$ between 0
and 1 with $f'(c)=0$. What is it?
Solution: To find $c$, we first compute $f'(x)= 3x^2  12x
+ 5$. So $$0=f'(c) = 3c^2  12 c + 5. $$ We then solve this with
the quadratic formula: $$c = \frac{12 \pm \sqrt{(12)^2 
4(3)(5)}}{6} = \frac{12 \pm \sqrt{84}}{6} = \frac{6 \pm
\sqrt{21}}{3}$$ Since we are looking for something between 0 and
1, we want $c = (6\sqrt{21})/3 \approx 0.4725$ instead of
$(6+\sqrt{21})/3 \approx 3.5275$. (If we had used Rolle's theorem
with $a=1$ and $b=5$, we would have wanted the larger value of
$c$.)
This answer isn't a particularly nice number. But nothing guarantees
that realworld problems will have simple answers!
Example 2: Consider the function $f(x) = e^x$ between $x=1$
and $x=3$. Find a point where the line tangent to the curve $y=f(x)$
is parallel to the line between $(1,e)$ and $(3,e^3)$. The
$x$value of this point is the $c$ in the MVT.
DO: Try to find the value
of $c$ before reading further.
Solution: This is the Mean Value Theorem with $f(x)=e^x$,
$a=1$ and $b=3$. We solve it in steps:
 We compute $\displaystyle \frac{f(b)f(a)}{ba} =
\frac{e^3e}{2}$.
 We look for c. Since $f'(x)=e^x$, ${\displaystyle f'(c) = e^c
= \frac{e^3e}{2}}$, so ${\displaystyle c = \ln \left (
\frac{e^3e}{2} \right ) \approx 2.1614}$.
 Once we have $c$, we have our point $$ (c, e^c) = \left (\ln
\left (\frac{e^3e}{2} \right ), \frac{e^3e}{2} \right) \approx
(2.1614, 8.684).$$
