Calculus I Lesson 16: Mean Value Theorem

  1. Warm Up
  2. Homework questions?
  3. The Mean Value Theorem and its applications
    1. Increasing / Decreasing / Constant
    2. First Derivative Test
    3. Second Derivative Test
    4. Concavity
  4. Upcoming / Reminders

Warm Up

Find the linear approximation (the equation of the tangent line) to $f(x) = 2\sin(2x)$ at $x = \frac{\pi}{2}$. Use this to approximate $f(1.5)$, using the approximation that $\frac{\pi}{2} \approx 1.57$.

Homework questions?

The Mean Value Theorem and its applications

The next major application of derivatives involves the shape of graphs. We have studied this already, but not really from a rigorous perspective. That is, intuitively we know that if $f^\prime(a) > 0$, then $f(x)$ is increasing at (or near) that point. But how do we know this? Can we justify this with a rigorous proof?

To actually prove this fact, we need the Mean Value Theorem:

Theorem: Let $f(x)$ be continuous over the interval $[a, b]$ and differentiable over $(a, b)$. Then there is a point $c$ in the interval $(a, b)$ such that \(f^\prime(c) = \frac{f(b) - f(a)}{b - a}\)

That is, there is a point inside the interval whose derivative is exactly equal to the average rate of change between $a$ and $b$.

We will not prove this theorem, but if you are interested, you can look at the proof in Section 4.4 of the textbook. I do want to point your attention to Figure 4.25:

A vaguely sinusoidal function y = f(x) is drawn. On the x-axis, a, c1, c2, and b are marked. On the y-axis, f(a) and f(b) are marked. The function f(x) starts at (a, f(a)), decreases to c1, increases to c2, and then decreases to (b, f(b)). A secant line is drawn between (a, f(a)) and (b, f(b)), and it is noted that this line has slope (f(b) – f(a))/(b − a). The tangent lines at c1 and c2 are drawn, and these lines are parallel to the secant line. It is noted that the slopes of these tangent lines are f’(c1) and f’(c2), respectively.

The mean value theorem says that there is a point $c$ where the tangent line is parallel to the secant line between $(a, f(a))$ and $(b, f(b))$. As an example, imagine that you are on a road trip. Let’s say that you were driving for 2 hours before stopping, and in those 2 hours, you drove 80 miles. That means, on average, you drove 40mph. The Mean Value Theorem says that there must be a time during that drive where your speedometer showed exactly 40mph!

Exercise: You pass a speed camera at 10:00 AM going 50 mph. At 10:30 AM you pass another speed camera located 35 miles apart, again going 50 mph. If the speed limit is 60 mph, were you ever speeding?

Take a look at exercises 190-191 in Section 4.4 of the textbook.

Increasing / Decreasing / Constant

The main application of the mean value theorem is to provide rigorous proofs of some of the intuitive notions we have about the shape of the graph of a function. Intuitively, if $f^\prime(x)$ is positive on an interval, then $f(x)$ should be increasing on that interval. How do we actually prove this? In other words, $f^\prime(x) > 0$ means that for each $x$, \(\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} > 0\) How do we then conclude that whenever $a < b$, then $f(a) < f(b)$?

We do so using the mean value theorem!

Theorem: Suppose $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$. If $f^\prime(x) > 0$ for every $x$ in $(a, b)$, then $f$ is an increasing function over $[a, b]$.

Proof: Suppose $x_0 < x_1$ are in the interval $[a, b]$. Then by the mean value theorem, there is a point $c$ between $x_0$ and $x_1$ such that \(f^\prime(c) = \frac{f(x_1) - f(x_0)}{x_1 - x_0}\). Since $f^\prime(c) > 0$ and $x_1 - x_0 > 0$, this means that $f(x_1) - f(x_0) > 0$. In other words, $f(x_1) > f(x_0)$ whenever $x_1 > x_0$ in the interval $[a, b]$.

Similarly, we can prove that if $f^\prime(x) < 0$ on an interval, then $f$ is decreasing over that interval, and if $f^\prime(x) = 0$ on an interval, then $f$ is constant on that interval. These proofs are found in Section 4.4 of the textbook.

This last result gives us an important property of derivatives: if two functions have the same derivative, they might not be the same function, but they must differ by a constant. For example, the function $f(x) = x^2$ and the function $g(x) = x^2 + 2$ have the same derivative, $2x$, but are not the same function.

In other words, two functions whose derivatives are the same would have to look “parallel” to each other:

Notice that at every point, the tangent lines have the same slope, and the distance between the graphs doesn’t change.

Theorem: Suppose $f^\prime(x) = g^\prime(x)$ for all x in $(a, b)$. Then $f(x) - g(x)$ is a constant.

Proof: Let $h(x) = f(x) - g(x)$. Then $h^\prime(x) = f^\prime(x) - g^\prime(x)$. Since, for every $x$ in $(a, b)$, $f^\prime(x) = g^\prime(x)$, then $h^\prime(x) = 0$ on this interval. Theorem 4.6 from the textbook implies then that $h(x)$ is a constant function on the interval $(a, b)$. Therefore $f(x) - g(x)$ is a constant.

Example: Consider the functions $f(x) = (\sec(x))^2$ and $g(x) = (\tan(x))^2$. Their derivatives are $f^\prime(x) = 2\sec(x)\sec(x)\tan(x)$, which simplifies to $2(\sec(x))^2\tan(x)$, and $g^\prime(x) = 2\tan(x)(\sec(x))^2$. Notice that $f^\prime(x) = g^\prime(x)$ for every $x$. But $f(0) = 1$ and $g(0) = 0$, so these are not the same function. But in general, $f(x) = g(x) + 1$:

First Derivative Test

Recall that if $f(x)$ has a local extremum (minimum or maximum) at a point $x = a$, then either $f^\prime(a) = 0$ or $f^\prime(a)$ does not exist. That is, $x = a$ must be a critical point of the function. But before, we saw that not every critical point of a function is a local extremum: for example, $f(x) = x^3$ does not have a minimum or maximum at $x = 0$.

So, if we find the critical points of a function, how can we tell if there is a maximum, minimum, or neither at those critical points? Take a look at this image (Figure 4.31) from Section 4.5 of the textbook:

A function f(x) is graphed. It starts in the second quadrant and increases to x = a, which is too sharp and hence f’(a) is undefined. In this section f’ > 0. Then, f decreases from x = a to x = b (so f’ < 0 here), before increasing at x = b. It is noted that f’(b) = 0. While increasing from x = b to x = c, f’ > 0. The function has an inversion point at c, and it is marked f’(c) = 0. The function increases some more to d (so f’ > 0), which is the global maximum. It is marked that f’(d) = 0. Then the function decreases and it is marked that f’ > 0.

Notice that when $f$ has a local maximum, then to the left $f$ is increasing, and to the right it’s decreasing. In terms of the derivative, that means $f^\prime > 0$ to the left of a local maximum, and $f^\prime < 0$ to the right. This makes sense intuitively: the graph increases up until the peak, and then decreases afterward.

Similarly, if $f^\prime < 0$ to the left of $x = a$ and then $f^\prime > 0$ to the right, then $x = a$ is a local minimum. What happens at those points where $f^\prime > 0$ on both sides, or when $f^\prime < 0$ on both sides? That’s when $f$ does not have a local maximum or a local minimum at that point.

Exercise: Find all the critical points of $f(x) = x^3 - 3x$. Classify each critical point as being a local maximum, local minimum, or neither.

Solution:

First take the derivative $f^\prime(x) = 3x^2 - 3$. Setting this equal to 0 and solving, we get $3(x^2 - 1) = 0$, or $(x+1)(x-1) = 0$. This means either $x = -1$ or $x = 1$.

So our critical points are $x = -1$ and $x = 1$. We need to determine if $f^\prime(x)$ is positive or negative to the left of $x = -1$, between $x = -1$ and $x = 1$, and to the right of $x = 1$.

Because $f^\prime(x)$ is continuous, we only need to check one point in each of the intervals $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$.

Let's check $x = -2$: $f^\prime(-2) = 3(4) - 3 = 9$, which is positive, so $f$ is increasing on $(-\infty, -1)$.

Check $x = 0$: $f^\prime(0) = 3(0) - 3 = -3$, which is negative, so $f$ is decreasing on $(-1, 1)$.

Check $x = 2$: $f^\prime(2) = 3(4) - 3 = 9$, which is postiive, so $f$ is increasing on $(1, \infty)$.

Therefore, $f$ increases until $x = -1$, decreases until $x = 1$, and then increases again off to infinity. So $x = -1$ is a local maximum and $x = 1$ is a local minimum

The textbook solves these in a slightly different way using “sign analysis”. I’ll let you check that out yourself: see Example 4.17 in Section 4.5.

We can use this first derivative test to study properties of the graph of a function, given the graph of its derivative. For example, if you see that the graph of $f^\prime$ is positive, then crosses the $x$-axis at the point $x = a$, and then goes negative, then $x = a$ is a local minimum. Take a look at exercises 201-207 and 221-222 from Section 4.5 to practice this.

Second Derivative Test

Last time we alluded to the fact that you can use the second derivative to classify maxima and minima. The first derivative test says:

then $f(c)$ is a local maximum. If $f$ is twice differentiable on that interval, then notice that $f^\prime$ is decreasing at $x = c$, since it goes from positive, to 0, to negative. Therefore, $f^{\prime\prime}(c) < 0$. This is the idea behind the Second Derivative Test:

If $f^\prime(c) = 0$ and $f^{\prime\prime}$ is continuous over an interval containing $c$, then:

Example: Consider the function $f(x) = x^4 - 2x^2$. Find its critical points and classify all of them as local minima, maxima, or neither.

Solution: Since $f^\prime(x) = 4x^3 - 4x$, we set this equal to 0 to find the critical points. Factoring out $4x$ we get $4x(x^2 - 1) = 0$. Divide by 4 and factor:

\[x(x+1)(x-1) = 0\]

so our critical points are $x = -1, x = 0, x = 1$.

The second derivative $f^{\prime\prime}(x) = 12x^2 - 4$. Plug in each of these:

Graph:

Concavity

Another important feature of the shape of the graph of a function is the concavity of the function. The textbook describes concavity as referring to whether a function “curves upward” or “curves downward”. Personally, I have trouble figuring out what “curving upward” or “curving downward” means just by looking at a graph. However, I’d like to point out the following graph:

Notice that at $x = -1$, the tangent line is drawn above the graph of $f(x)$, and at $x = 1.5$, the tangent line is below the graph. This is how I understand concavity: the graph is concave up at a point (or “curves upward”) if it lies above the tangent line at that point, and is concave down (or “curves downward”) if it lies below the tangent line.

It turns out that concavity of a function is determined by whether the first derivative is increasing or decreasing: if $f^\prime$ is increasing on an interval, then $f$ is concave up, and if $f^\prime$ is decreasing, then $f$ is concave down. Of course, if $f$ has a second derivative, then we can just check if that second derivative is positive or negative: if $f^{\prime\prime} > 0$ on an interval, then $f$ is concave up on that interval. If $f^{\prime\prime} < 0$ on an interval, then $f$ is concave down on that interval.

An inflection point is a point where $f$ changes concavity. Again, if the function has a second derivative, we can check where $f^{\prime\prime}$ goes from positive to negative, or negative to positive: this would occur either at points where $f^{\prime\prime} = 0$ or where the second derivative is not defined.

Exercise: For the function $f(x) = x^4 - 4x^3 - 18x^2 - x + 3$, determine all intervals where $f$ is concave up, all intervals where $f$ is concave down, and all points of inflection. Sketch a graph and locate these intervals on the graph.

Check your solution

Let's check the second derivative. $f^\prime(x) = 4x^3 - 12x^2 - 36x - 1$, and so $f^{\prime\prime}(x) = 12x^2 - 24x -36$. Since this is continuous, we check where it is equal to 0:

$12x^2 - 24x - 36 = 0$. Factor out 12: $12(x^2 - 2x - 3) = 0$. Divide by 12 and factor.

$(x-3)(x+1) = 0$, and so the second derivative is equal to 0 at $x = -1$ and $x = 3$.

For each of the intervals $(-\infty, -1)$, $(-1, 3)$ and $(3, \infty)$, we check whether $f^{\prime\prime}$ is positive or negative:

For $(-\infty, -1)$, check the points $x = -2$: $f^{\prime\prime}(-2) = 12(4) - 24(-2) - 36 = 60 > 0$, so $f$ is concave up on $(-\infty, -1)$.

For $(-1, 3)$, check $x = 0$: $f^{\prime\prime}(0) = -36 < 0$, so $f$ is concave down on $(-1, 3)$. At this point we can conclude that $x = -1$ is an inflection point, since $f$ changes concavity there.

For $(3, \infty)$, check $x = 4$: $f^{\prime\prime}(4) = 60 > 0$, so $f$ is concave up on $(3, \infty)$. Again, $x = 3$ is an inflection point. Take a look at the graph and sketch out the tangent lines to see where the graph is concave up and concave down:

Take a look at exercises 211-217 and 224-230 from Section 4.5.

Upcoming / Reminders