Calculus/Some Important Theorems/Solutions

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Rolle's Thoerem[edit]

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 Theorem[edit]

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].

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.