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 $c$.)

1. We compute $\displaystyle \frac{f(b)-f(a)}{b-a} = \frac{e^3-e}{2}$.
2. We look for c. Since $f'(x)=e^x$, ${\displaystyle f'(c) = e^c = \frac{e^3-e}{2}}$, so ${\displaystyle c = \ln \left ( \frac{e^3-e}{2} \right ) \approx 2.1614}$.
3. Once we have $c$, we have our point $$(c, e^c) = \left (\ln \left (\frac{e^3-e}{2} \right ), \frac{e^3-e}{2} \right) \approx (2.1614, 8.684).$$