# Rolle's ThoeremEdit

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

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

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

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

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

Thus, at ${\displaystyle x=3/2}$, we have a spot with a slope of zero. We know that ${\displaystyle 3/2}$ (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 ${\displaystyle h(a)=h(b)}$, where ${\displaystyle h(x)}$ is the function that was defined in the proof of Cauchy's Mean Value Theorem.

{\displaystyle {\begin{aligned}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{aligned}}} {\displaystyle {\begin{aligned}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{aligned}}}

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

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

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

1: Using the expression from the mean value theorem

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

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

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

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 ${\displaystyle x=c}$.

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

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

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

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

so,

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

(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!

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

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