# Section6.3Consequences of the Mean Value Theorem¶ permalink

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.

# Subsection6.3.1Describing 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 $f'(x)$.

2. Solve the equation $f'(x)=0$.

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.

##### Example6.3.2

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

Solution
##### Example6.3.3

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

Solution

# Subsection6.3.2Describing 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.

##### Definition6.3.4Critical Points

Given a function $f$ with domain $D$, 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.

##### Example6.3.6

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

Solution

# Subsection6.3.3Describing 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.

##### Definition6.3.7Concavity

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

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 $f$, 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*}

##### Example6.3.9

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

##### Definition6.3.10Inflection 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 $x=a$. 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.

##### Example6.3.11

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

Solution

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 $x=a$. Consequently, $f'(x) \lt 0$ in an interval with $x \lt a$ and $f'(x) \gt 0$ in an interval with $x \gt 0$. By the First Derivative Test, $f(x)$ must have a minimum at $x=a$.

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 $f''(a) \gt 0$, 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 $f''(a)=0$. In such a case, we are forced to go back and use the First Derivative Test to see if monotonicity changes.

##### Example6.3.13

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

Solution

From a practical point of view, in order to find the critical points, you will already have had to factor $f'(x)$. 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 $f'(x)$.