Tag Archives: Mean Value Theorem

Rolle’s Theorem

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Rolle’s theorem says that if a function is continuous on a closed interval [a, b], differentiable on the open interval (a, b) and if f (a) = f (b), then there exists a number c in the open interval (a, b) such that {f}'\left( c \right)=0.  (“There exists a number” means that there is at least one such number; there may be more than one.)

The proof has two cases:

Case I: The function is constant (all of the values of the function are the same as f (a) and f (b)). The derivative of a constant is zero so any (every, all) value(s) in the open interval qualifies as c.

Case II: If the function is not constant then it must have a maximum or minimum in the open interval (a, b) by the Extreme Value Theorem. So, by Fermat’s theorem (see this post) the derivative at that point must be zero.

So, Fermat’s theorem makes Rolle’s theorem a piece of cake.

A lemma is a theorem whose result is used in the next theorem and makes it easier to prove. So Fermat’s theorem is a lemma for Rolle’s theorem.

On the other hand, a corollary is a theorem is a result (theorem) that follows easily from the previous theorem. So, Rolle’s theorem could also be called a corollary of Fremat’s theorem.

Rolle’s theorem makes a major appearance in the MVT and then more or less disappears from the stage. When you find critical number or critical points you are using Fermat’s theorem.

I like this proof because it’s so simple. It really just comes immediately from Fermat’s theorem.

The next post: The Mean Value Theorem.

Fermat’s Penultimate Theorem

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I have mixed feelings about proof in high school math and high school calculus. I am not one for proving everything. For one thing, it cannot be done and, if it could be done, proof would become the whole focus of high school math. Proofs are not the focus of first-year calculus or AP calculus. The place for proving “everything” is a real analysis course in college.

However, students should know about proof and there are places where you can demonstrate some of the power of proof and show how proof works in calculus.
It is important, I think, that students know why a theorem is true; this helps in understanding what the theorem means. Some, but by no means all, proofs can show the student why the theorem is true. With other theorems there may be easier ways than a proof to convince someone of its truth.

In this and the next three posts, I propose to look at three theorems, the definitions used in them, and the ideas in their proofs. These are the theorems that lead up to the Mean Value Theorem (MVT). The MVT is a major result in calculus has many uses. Here goes:

Fermat’s theorem (not his famous “last” theorem, but an earlier one) says, that if a function is continuous on a closed interval and has a maximum (or minimum) value on that interval at x = c, then the derivative at x = c is either zero or does not exist.

The proof goes like this:
There are two cases. In each case we will look at the limit of the difference quotient that defines the derivative at x = c, namely, \frac{f\left( c+h \right)-f\left( c \right)}{h} and look at what happens as h approaches 0 from the left and from the right. These two limits are the same and equal to the derivative if, and only if, the derivative at c exists.

Also note that since we are assuming f(c) is a maximum, f (c) ≥ f (c + h) regardless of whether h is positive or negative. The numerator of the difference quotient is always zero or negative. Then if in the denominator h < 0, the quotient is non-positive; likewise, if h > 0, the quotient is non-negative.

Case I: The two limits are not equal. In this case the derivative does not exist. This could occur with a piecewise function, where two pieces with different derivatives meet at x = c.

Case II: The limits are equal. In this case the limit from the left (h < 0) must be greater than or equal to zero (since the function is increasing there) and the limit from the right (h > 0) must be less than or equal to zero. Then, the only way the limits can be equal is if both limits are zero; therefore the derivative is zero.

Any place where the derivative of a continuous function is zero or undefined is called a critical point and the number c is called a critical number (new definitions).

I think this proof is interesting because while there are lots of symbols flying around the key is interpreting what kind of number (positive, zero or negative) the symbols represent. Another thing I like is having to “read” the symbols and see that f\left( c+h \right)\le f\left( c \right) and therefore f\left( c+h \right)-f\left( c \right)\le 0

The next post will discuss Rolle’s theorem.

Local Linearity I

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Certain graphs, specifically those that are differentiable, have a property called local linearity. This means that if you zoom in (using the same zoom factor in both directions) on a point on the graph, the graph eventually appears to be a straight line whose slope if the same as the slope (derivative) of the tangent line at that point.

Now we are a little ahead of ourselves here since we haven’t mentioned tangent lines and derivatives yet. But local linearity is the graphical manifestation of differentiability. Functions that are differentiable at a point are locally linear there and functions that are locally linear are differentiable. In the next post we will see how to use the local linear idea to introduce the derivative. For now, we will look at some graphs that may or may not be locally linear. Are the graphs “smooth” everywhere?

(1)\quad f\left( x \right)=1+\sqrt{{{x}^{2}}+0.001}

\displaystyle (2)\quad g\left( x \right)={{x}^{3}}+\frac{\sqrt[3]{{{\left( x-1 \right)}^{2}}}}{7}

The first function is locally linear at (0, 1) but doesn’t look it. Zoom-in several times and you will see that it is smooth and locally linear there eventually the graph looks like a horizontal line near (0, 1). This only looks like an absolute value graph because 1+\sqrt{{{x}^{2}}+0.001}\approx 1+\sqrt{{{x}^{2}}}=1+\left| x \right|.

The second function appears locally linear at (1, 1), but is not. Zoom in a few times at you will see very strange things going on. (Hint: Use a graphing program or a calculator and enter x^3+((x – 1)^2)^(1/3)/7 as simplifying to a power of 2/3 may confuse the calculator.)

The moral is that you can never be sure just looking at a graph whether it is locally linear or not; you’re never sure if you have zoomed in enough.

Nevertheless, the local linearity concept is helpful in introducing the derivative and, when you can be sure the function is not locally linear, knowing the derivative does not exist.

Looking ahead: Take a look at the AP exam from 2005 AB 5. Here you were given a velocity graph “modeled by the piecewise- linear function defined by the graph” copied below.

Student were asked to find v ‘(4) or explain why it does not exist. It does not exist because the graph is not locally linear there.

Later they were asked if the Mean Value Theorem guaranteed a value of c in the interval [8, 20] such that v ‘(c) is equal to the average rate of change over the interval. The answer is “no”, the value does not exist because the function is not differentiable on the interval because it is not locally linear everywhere in the interval.