Calculus/Some Important Theorems/Solutions

Rolle's Thoerem

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

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.

3. Show that the Mean Value Theorem follows from Cauchy's 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 $[0,\pi ]$.

1: We start with the expression:

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

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.