# Rolle's Thoerem

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).
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 Theorem

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{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}}  {\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}}
{\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}}  {\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 $g(x)=x$ . Then $g'(x)=1$  and $g(b)-g(a)=b-a$ , which is non-zero if $b\neq a$ . 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.
Let $g(x)=x$ . Then $g'(x)=1$  and $g(b)-g(a)=b-a$ , which is non-zero if $b\neq a$ . 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 $x={\sqrt {4/3}}=\mathbf {\frac {2{\sqrt {3}}}{3}}$
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.
${\frac {f(b)-f(a)}{b-a}}$

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.