## Section9.2Consequences of the Mean Value Theorem

Just as the Mean Value Theorem for Integrals was used to describe the monotonicity of accumulation functions in terms of the sign of the rate function, the Mean Value Theorem allows us to find the monotonicity of any differentiable function in terms of the sign of its derivative.

Consider any two points $a,b \in I$ with $a \lt b\text{.}$ Because $f$ is differentiable on $I\text{,}$ we know that $f$ is continuous and differentiable on the subinterval $[a,b]\text{.}$ The Mean Value Theorem guarantees the existence of a point $c \in (a,b)$ such that

\begin{equation*} f(b)-f(a) = f'(c)\cdot(b-a). \end{equation*}

Now assume that $f'(x) \gt 0$ for all $x \in I\text{.}$ Then $f'(c) \gt 0$ and $b-a \gt 0\text{,}$ guaranteeing that $f(b)-f(a) \gt 0\text{.}$ That is, $f(b) \gt f(a)\text{.}$ This is what is needed to show that $f$ is increasing on $I\text{.}$

Next assume that $f'(x) \lt 0$ for all $x \in I\text{.}$ Then $f'(c) \lt 0$ while $b-a \gt 0\text{,}$ guaranteeing that $f(b)-f(a) \lt 0\text{.}$ That is, $f(b) \lt f(a)\text{,}$ which shows that $f$ is decreasing on $I\text{.}$

Finally assume that $f'(x) = 0$ for all $x \in I\text{.}$ Then $f'(c) = 0\text{,}$ implying that $f(b)-f(a) = 0\text{.}$ That is, $f(b) = f(a)\text{,}$ which shows that $f$ is constant on $I\text{.}$

### Subsection9.2.1Describing Monotonicity Using Derivatives

Because of the theorem about monotonicity, we can understand the behavior of a function if we can find all of the intervals on which the derivative is positive, negative or zero. The only points where $f'$ might change sign are where $f'(x)=0$ or where $f'(x)$ is discontinuous.

Our strategy will be as follows.

1. Compute $f'(x)\text{.}$

2. Solve the equation $f'(x)=0\text{.}$

3. Use the roots and discontinuities of $f'(x)$ to identify all relevant intervals.

4. Test the sign of $f'(x)$ on each interval.

5. Apply the Monotonicity theorem to determine the behavior of $f$ on each interval.

###### Example9.2.2

Describe the intervals of monotonicity for the function $f(x) = x^3-2x^2-4x\text{.}$

Solution
• Compute $f'(x) = 3x^2-4x-4\text{.}$

• Solve $f'(x)=0\text{.}$ This will require factoring.

\begin{gather*} 3x^2-4x-4=0\\ (3x+2)(x-2)=0 \end{gather*}

There are two possibilities, $3x+2=0$ or $x-2=0\text{.}$ The first leads to $x=-\frac{2}{3}\text{;}$ the second leads to $x=2\text{.}$

• $f'(x)$ exists for every value $x\text{,}$ so we use these two roots to break the number line into intervals: $(-\infty, -\frac{2}{3})\text{,}$ $(-\frac{2}{3},2)$ and $(2,\infty)\text{.}$ It can be convenient to use a graphical number line to identify the intervals, as shown below. We put $x$-values below the line and the corresponding values of $f'(x)$ above the line.

• We test the sign of $f'(x)$ at a point in each of the intervals. Because we factored $f'(x) = (3x+2)(x-2)\text{,}$ we can quickly determine the sign by looking at the signs of the factors. We do not need the actual value, so I will summarize the sign of the factor with $+$ or $-\text{.}$ We need to choose one point from each interval.

\begin{align*} f'(-1) &= (3(-1)+2)(-1-2) = (-)(-) = +\\ f'(0) &= (3(0)+2)(0-2) = (+)(-) = -\\ f'(3) &= (3(3)+2)(3-2) = (+)(+) = + \end{align*}

We can summarize these results on our number line by marking the sign of $f'(x)$ above each interval.

• We finish by interpreting the results. Because $f(x)$ is a polynomial, it is continuous. We can extend all of our monotonicity conclusions to include the end-points.

$f'(x) \gt 0$ on $(-\infty,-\frac{2}{3})$ $\implies$ $f(x)$ is increasing on $(-\infty,-\frac{2}{3}]\text{.}$

$f'(x) \lt 0$ on $(-\frac{2}{3},2)$ $\implies$ $f(x)$ is decreasing on $[-\frac{2}{3}, 2]\text{.}$

$f'(x) \gt 0$ on $(2,\infty)$ $\implies$ $f(x)$ is increasing on $[2,\infty)\text{.}$

A graph of $y=f(x)$ is shown below for comparison.

###### Example9.2.3

Describe the intervals of monotonicity for the function $f(x) = xe^{x-2x^2}\text{.}$

Solution
• Compute $f'(x)=\frac{d}{dx}[xe^{x-2x^2}]\text{,}$ which requires a product rule:

\begin{align*} f'(x) &= x \frac{d}{dx}[e^{x-2x^2}] + e^{x-2x^2}\frac{d}{dx}[x]\\ &= x \cdot e^{x-2x^2} \frac{d}{dx}[x-2x^2] + e^{x-2x^2} \cdot 1 \\ &= x(1-4x)e^{x-2x^2} + e^{x-2x^2} \end{align*}
• Solve the equation $f'(x)=0\text{.}$ We need to simplify and then factor $f'(x)\text{.}$

\begin{align*} f'(x) &= (x-4x^2) e^{x-2x^2} + e^{x-2x^2}\\ &= e^{x-2x^2} \big( x-4x^2 + 1 \big) \end{align*}

Now solve the equation: $e^{x-2x^2} (-4x^2+x+1) = 0\text{.}$ Because $e^x \gt 0$ for any $x\text{,}$ $e^{x-2x^2}=0$ has no solutions. The only possible solutions come from $-4x^2+x+1 = 0\text{.}$ This is quadratic $ax^2+bx+c=0$ so that solutions are

\begin{align*} x &= \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-1 \pm \sqrt{1-4(-4)(1)}}{2(-4)} \\ &= \frac{-1 \pm \sqrt{17}}{-8} \end{align*}

These roots are $x=-\frac{1}{8}-\frac{\sqrt{17}}{8} \approx -0.390$ and $=-\frac{1}{8}+\frac{\sqrt{17}}{8}\approx 0.640\text{.}$

• Since $f'(x)$ is defined for all $x\text{,}$ we use the roots to identify intervals.

Testing points in each of the intervals, at $x \in \{-1,0,1\}\text{,}$ we can find the sign of $f'(x)=e^{x-2x^2}(-4x^2+x+1)\text{.}$

$f'(-1) = e^{-1-2} (-4(-1)^2-1+1) = (+)(-) = -$

$f'(0) = e^{0} (-4(0)^2+0+1) = (+)(+) = +$

$f'(1) = e^{1-2} (-4(1)^2+1+1) = (+)(-) = -$

Updating the number line with this information gives us the following figure.

• We interpret our results about each interval.

$f'(x) \lt 0$ on $(-\infty, \frac{1-\sqrt{17}}{8})$ $\implies$ $f(x)$ is decreasing on $(-\infty, \frac{1-\sqrt{17}}{8}]\text{.}$

$f'(x) \gt 0$ on $(\frac{1-\sqrt{17}}{8},\frac{1+\sqrt{17}}{8})$ $\implies$ $f(x)$ is increasing on $[\frac{1-\sqrt{17}}{8}, \frac{1+\sqrt{17}}{8}]\text{.}$

$f'(x) \lt 0$ on $(\frac{1+\sqrt{17}}{8},\infty)$ $\implies$ $f(x)$ is decreasing on $[\frac{1+\sqrt{17}}{8},\infty)\text{.}$

A graph of $y=f(x)$ is shown below for comparison.

### Subsection9.2.2Describing Extreme Values Using Derivatives

Fermat's theorem states that a function $f$ can have an extreme value only at points where either $f'(x)=0$ or $f'(x)$ does not exist. This motivates the definition of critical points.

###### Definition9.2.4Critical Points

Given a function $f$ with domain $D\text{,}$ a value $c \in D$ is a critical point of $f$ if $f'(c)=0$ or $f'(c)$ does not exist.

Not every critical point is an extreme value. We need criteria to judge whether or not a given critical point is the location of a maximum or minimum value. Intervals of monotonicity, as determined by the derivative, allow us to make that conclusion. Using monotonicity to classify critical points is justified by the following theorem, known as the first derivative test.

The first derivative test is usually applied by looking at the number line summary of the first derivative.

###### Example9.2.6

Find and classify the local extremes of $f(x) = x^3(x^2-4)\text{.}$

Solution

Start by finding the critical points, which requires knowing the derivative.

Now solve $f'(x)=0\text{,}$ which gives either $x^2=0$ or $5x^2-12=0\text{.}$ There are three solutions: $x=0$ and $x=\pm \sqrt{\frac{12}{5}}\text{.}$ Because $f'(x)$ is defined for all values of $x\text{,}$ these three are the only critical points.

We test if the critical points are local extremes by seeing if the direction of monotonicity changes at these points. This is summarized using a number line. Testing the value of $f'(x)$ at points, such as the set $\{-2, -1, 1, 2\}\text{,}$ gives us the following signs for the rate of change.

We interpret our results by applying the First Derivative Test. The critical points at $x=\pm\sqrt{\frac{12}{5}}$ are turning points. But $x=0$ is not a turning point even though $f'(0)=0\text{.}$ Instead, it is a point of inflection. (We would need to show that $f''(x)$ changes sign at $x=0\text{.}$) The point $x=-\sqrt{\frac{12}{5}}$ is a local maximum because it is a turning point going from increasing to decreasing. The point $x=+\sqrt{\frac{12}{5}}$ is a local minimum because it is a turning point going from decreasing to increasing. A graph of $y=f(x)$ is shown below.

### Subsection9.2.3Describing Concavity Using Derivatives

Concavity describes changing slope or rate of change. On a graph, this corresponds to how the curve described by a function is bending. We restate the definition of concavity in terms of the rate of change defined by a derivative.

###### Definition9.2.7Concavity

Let $f$ be a function such that the derivative $f'$ is defined on an interval $I\text{.}$ We say $f$ is concave up on $I$ if $f'$ is increasing on $I\text{.}$ We say $f$ is concave down on $I$ if $f'$ is decreasing on $I\text{.}$

The four quadrants of a circle illustrate the four possible combinations of monotonicity and concavity.

increasing, concave down (II)

The function has a positive rate of change and that rate of change is decreasing. The graph has a positive slope that is becoming less steep (less positive).

increasing, concave up (IV)

The function has a positive rate of change and that rate of change is increasing. The graph has a positive slope that is becoming steeper (more positive).

decreasing, concave down (I)

The function has a negative rate of change and that rate of change is decreasing. The graph has a negative slope that is becoming steeper (more negative).

decreasing, concave up (III)

The function has a negative rate of change and that rate of change is increasing. The graph has a negative slope that is becoming less steep (less negative).

Concavity is described by finding where the derivative is increasing or decreasing. Since the derivative is itself a function, we can find its derivative. This function will be the derivative of the derivative of the function $f\text{,}$ or the second derivative. We write

\begin{equation*} f''(x) = \frac{d}{dx}[f'(x)] = \frac{d^2}{dx^2}[f(x)] = \frac{d^2f}{dx^2}. \end{equation*}

This is just an application of Theorem Theorem 9.2.1 applied to $f'\text{.}$ Once we know that $f'$ is increasing on $I\text{,}$ the definition of concavity allows us to say that $f$ is concave up on $I\text{.}$ Similarly, knowing that $f'$ is decreasing is equivalent to saying that $f$ is concave down. If $f'$ is constant on an interval $I\text{,}$ this is exactly what it means for $f$ to be linear on $I\text{.}$

###### Example9.2.9

The concavity of a parabola is completely determined by the sign of the leading coefficient. Suppose $f(x) = ax^2+bx+c\text{,}$ where $a\text{,}$ $b$ and $c$ are parameters. Then $f'(x) = 2ax+b$ and $f''(x)=2a\text{.}$ The second derivative for a quadratic is constant, and the sign is determined by the sign of $a\text{.}$ If $a \gt 0\text{,}$ $f$ is concave up on $(-\infty,\infty)\text{;}$ if $a \lt 0\text{,}$ $f$ is concave down on $(-\infty,\infty)\text{.}$

###### Definition9.2.10Inflection Point

A differentiable function $f$ has an inflection point at a point $x=a$ if $a \in \mathrm{dom}(f)$ and $f$ changes concavity at $x=a\text{.}$ That is, there is are intervals $(a-\delta,a)$ and $(a,a+\delta)$ for $\delta \gt 0$ such that $f$ has opposite concavity in these intervals.

###### Example9.2.11

Given $f(x) = x^3-3x^2-4x+5\text{,}$ describe where $f$ is concave up and concave down.

Solution

Start by finding $f'(x)$ and $f''(x)\text{.}$

\begin{align*} f(x) &= x^3-3x^2-4x+5 \\ f'(x) &= 3x^2-6x-4 \\ f''(x) &= 6x-6 \end{align*}

We now see where $f''(x)$ changes sign by solving $6x-6=0\text{.}$ This has a single solution $x=1\text{.}$ We test the sign of $f''(x)$ by choosing values for $x$ on either side of $x=1\text{.}$

\begin{gather*} f''(0) = -6 \\ f''(2) = 6 \end{gather*}

The signs of $f''(x)$ are summarized with a number line.

We interpret our results using Theorem Theorem 9.2.8. Because $f''(x) \lt 0$ on $(-\infty,1)\text{,}$ we know $f'$ is decreasing on the interval. Since $f'(x)$ is continuous, we can extend the interval to include the end-point and conclude that $f$ is concave down on $(-\infty,1]\text{.}$ Similarly, since $f''(x) \gt 0$ on $(1,\infty)$ and $f'$ is continuous at $1\text{,}$ we know $f$ is concave up on $[1,\infty)\text{.}$ Because concavity changes at $x=1\text{,}$ $f$ has an inflection point at $x=1\text{.}$

When a second derivative can be determined conveniently, there is another way to determine whether a critical point is a turning point corresponding to a local extreme. Suppose $f'(a)=0$ but $f''(a)$ is positive. This means that $f'(x)$ is increasing in a neighborhood of $x=a\text{.}$ Consequently, $f'(x) \lt 0$ in an interval with $x \lt a$ and $f'(x) \gt 0$ in an interval with $x \gt 0\text{.}$ By the First Derivative Test, $f(x)$ must have a minimum at $x=a\text{.}$

Graphically, this means that at $x=a$ there is a horizontal tangent with the graph concave up (like a parabola). This point then behaves like the vertex of a parabola opening upward and $x=a$ is a local minimum. Similarly if $f''(a) \gt 0\text{,}$ the function is concave down and the graph behaves somewhat like a downward opening parabola and the point $x=a$ would be a local maximum. These results are summarized as a theorem.

Notice that the Second Derivative Test is inconclusive when $f''(a)=0\text{.}$ In such a case, we are forced to go back and use the First Derivative Test to see if monotonicity changes.

###### Example9.2.13

Find and classify the local extremes of $f(x) = x^3(x^2-4)$ using the Second Derivative Test.

Solution

In the previous example, we solved the same problem. We found $f'(x) = 5x^4-12x^2\text{.}$ The critical points were at $x=0$ and $x=\pm \sqrt{\frac{12}{5}}\text{.}$ The Second Derivative Test has us compute $f''(x)=20x^3-24x$ and determine the sign of $f''(x)$ at each of the critical points.

\begin{align*} f''(x) &= 20x^3-24x = 4x(5x^2-6)\\ f''(0) &= 0\\ f''(-\sqrt{\frac{12}{5}}) &= -4 \sqrt{\frac{12}{5}} (5(\frac{12}{5})-6) = -24 \sqrt{\frac{12}{5}} \lt 0\\ f''(+\sqrt{\frac{12}{5}}) &= +4 \sqrt{\frac{12}{5}} (5(\frac{12}{5})-6) = 24 \sqrt{\frac{12}{5}} \gt 0 \end{align*}

The Second Derivative Test is inconclusive about $x=0$ being a turning point. Since $f(x)$ is locally concave down at the critical point $x=-\sqrt{\frac{12}{5}}\text{,}$ the Second Derivative Test implies that $f(x)$ has a local maximum at this point. Similarly, $f(x)$ has a local minimum at $x=+\sqrt{\frac{12}{5}}\text{.}$

From a practical point of view, in order to find the critical points, you will already have had to factor $f'(x)\text{.}$ If finding the signs of $f'(x)$ on the resulting intervals will be faster than computing $f''(x)$ and then computing the sign at each critical points, then the First Derivative Test is a better strategy. The second derivative test is useful if finding $f''(x)$ will be especially easy, or at least easier than finding the signs of $f'(x)\text{.}$