# Rolle's ThoeremEdit

1. Show that Rolle's Theorem holds true between the x-intercepts of the function $f(x)=x^2-3x$.

1: The question wishes for us to use the $x$-intercepts as the endpoints of our interval.

Factor the expression to obtain $x(x-3)= 0$. $x=0$ and $x=3$ are our two endpoints. We know that $f(0)$ and $f(3)$ are the same, thus that satisfies the first part of Rolle's theorem ($f(a)=f(b)$).

2: Now by Rolle's Theorem, we know that somewhere between these points, the slope will be zero. Where? Easy: Take the derivative.

$\frac{dy}{dx}$ $= 2x - 3$

Thus, at $x = 3/2$, we have a spot with a slope of zero. We know that $3/2$ (or 1.5) is between 0 and 3. Thus, Rolle's Theorem is true for this (as it is for all cases).

# Mean Value TheoremEdit

2. Show that $h(a)=h(b)$, where $h(x)$ is the function that was defined in the proof of Cauchy's Mean Value Theorem.

\begin{align}h(a)&=f(a)(g(b)-g(a)-g(a)(f(b)-f(a)-f(a)g(b)+f(b)g(a)\\ &=f(a)g(b)-f(a)g(a)-g(a)f(b)-g(a)f(a)-f(a)g(b)+f(b)g(a)\\ &=0\end{align} \begin{align}h(b)&=f(b)(g(b)-g(a))-g(b)(f(b)-f(a))-f(a)g(b)+f(b)g(a)\\ &=f(b)g(b)-f(b)g(a)-g(b)f(b)-g(b)f(a)-f(a)g(b)+f(b)g(a)\\ &=0\end{align}

3. Show that the Mean Value Theorem follows from Cauchy's Mean Value Theorem.

Let $g(x)=x$. Then $g'(x)=1$ and $g(b)-g(a)=b-a$, which is non-zero if $b\ne a$. Then
$\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}$ simplifies to $f'(c) = \frac{f(b) - f(a)}{b-a}$, which is the Mean Value Theorem.

4. Find the $x=c$ that satisfies the Mean Value Theorem for the function $f(x)=x^3$ with endpoints $x=0$ and $x=2$.

1: Using the expression from the mean value theorem

$\frac{f(b)-f(a)}{b-a}$

insert values. Our chosen interval is $[0,2]$. So, we have

$\frac{f(2)-f(0)}{2-0} = \frac{8}{2} = 4$

2: By the Mean Value Theorem, we know that somewhere in the interval exists a point that has the same slope as that point. Thus, let us take the derivative to find this point $x = c$.

$\frac{dy}{dx} = 3x^2$

Now, we know that the slope of the point is 4. So, the derivative at this point $c$ is 4. Thus, $4 = 3x^2$. So $x=\sqrt{4/3}=\mathbf{\frac{2\sqrt{3}}{3}}$

5. Find the point that satisifies the mean value theorem on the function $f(x) = \sin(x)$ and the interval $[0,\pi]$.

$\frac{f(b)-f(a)}{b-a}$

so,

$\frac{\sin(\pi) - \sin(0)}{\pi - 0} = 0$

(Remember, sin(π) and sin(0) are both 0.)

2: Now that we have the slope of the line, we must find the point x = c that has the same slope. We must now get the derivative!

$\frac{d\sin(x)}{dx} = \cos(x) = 0$

The cosine function is 0 at $\pi /2 + \pi n$ (where $n$ is an integer). Remember, we are bound by the interval $[0,\pi]$, so $\mathbf{\pi/2}$ is the point $c$ that satisfies the Mean Value Theorem.