Chapter 15: Increasing and Decreasing Functions and Local Extrema
Elementary
The derivative can be used to determine the intervals on which a function increases or decreases, as well as the points where it has local maxima and minima. In this chapter, we will learn how to use the sign of the derivative and how to find absolute maxima and minima with a sign chart.
15.1 Determining Increase and Decrease
Whether a function is increasing or decreasing can be determined from the sign of its derivative, $f'(x)$. This is because the derivative gives the slope of the tangent line at each point.
Suppose that $f$ is differentiable at every point in the interior of an interval $I$. If one of the following conditions holds throughout the interior of $I$, then:
- $f'(x) > 0$ ⇒ $f$ is increasing on $I$.
- $f'(x) < 0$ ⇒ $f$ is decreasing on $I$.
- $f'(x) = 0$ ⇒ $f$ is constant on $I$.
In other words, a positive derivative means that the graph rises from left to right, while a negative derivative means that it falls. At a point where the derivative is $0$, the tangent is horizontal, but that point is not necessarily a local maximum or minimum.
15.2 Local Extrema: Definitions and Tests
Let $a$ be a point in the domain of a function $f$. If one of the following inequalities holds for all $x$ sufficiently close to $a$, then $f(a)$ is called a local extremum:
- Local maximum: $f(x) \leq f(a)$ for all $x$ sufficiently close to $a$.
- Local minimum: $f(x) \geq f(a)$ for all $x$ sufficiently close to $a$.
A local maximum or minimum is the greatest or least value compared with nearby values. It should be distinguished from an absolute maximum or minimum, which is the greatest or least value over the entire domain.
Suppose that $f$ is continuous at $a$ and differentiable on both sides of $a$ in a neighborhood of that point. Examine the sign of $f'(x)$ to the left and right of $a$. The derivative does not need to be defined at $a$ itself.
- $f'(x) > 0$ to the left and $f'(x) < 0$ to the right ⇒ $f$ has a local maximum at $a$.
- $f'(x) < 0$ to the left and $f'(x) > 0$ to the right ⇒ $f$ has a local minimum at $a$.
- No sign change across $a$ ⇒ $f$ has neither a local maximum nor a local minimum at $a$.
This test applies not only when $f'(a)=0$, but also when $f'(a)$ does not exist. However, $a$ must belong to the domain of the function.
Suppose that $f'(a)=0$ and that $f$ is twice differentiable near $a$. The sign of $f''(a)$ can then be used to determine the type of local extremum:
- $f''(a) > 0$ ⇒ $f$ has a local minimum at $a$.
- $f''(a) < 0$ ⇒ $f$ has a local maximum at $a$.
- $f''(a) = 0$ ⇒ this test is inconclusive.
Even when $f''(a)=0$, the function may or may not have a local extremum. For example, $f(x)=x^4$ has a local minimum at $x=0$, whereas $f(x)=x^3$ has no local extremum there.
15.3 Constructing a Sign Chart
Step 1Find the derivative.
$f'(x) = 3x^2 - 3 = 3(x+1)(x-1)$
Step 2Solve $f'(x) = 0$.
$x = -1, 1$
Step 3Determine the sign of $f'(x)$ on each interval.
We have $f'(x)>0$ for $x < -1$ and $x > 1$, and $f'(x)<0$ for $-1 < x < 1$.
Also, $f(-1)=4$ and $f(1)=0$. These results can be summarized in the following sign chart.
| $x$ | $\cdots$ | $-1$ | $\cdots$ | $1$ | $\cdots$ |
|---|---|---|---|---|---|
| $f'(x)$ | $+$ | $0$ | $-$ | $0$ | $+$ |
| $f(x)$ | ↗ | Local maximum $4$ |
↘ | Local minimum $0$ |
↗ |
15.4 Absolute Maximum and Minimum Problems
If $f$ is continuous on the closed interval $[a,b]$, it attains both an absolute maximum and an absolute minimum on that interval. The candidates are:
- The endpoints, $x=a$ and $x=b$.
- Interior points where $f'(x)=0$.
- Interior points where $f'(x)$ does not exist (provided that $f(x)$ is defined there).
Evaluate the function at every candidate point. The largest value is the absolute maximum, and the smallest is the absolute minimum. Checking only local maxima and minima is not always enough to determine the absolute extrema on the entire interval.
Solving $f'(x) = 3x^2 - 3 = 0$ gives $x = \pm 1$.
Values at the candidate points:
- $f(-2) = -8 + 6 = -2$
- $f(-1) = -1 + 3 = 2$(local maximum)
- $f(1) = 1 - 3 = -2$(local minimum)
- $f(3) = 27 - 9 = 18$
Absolute maximum: $18$ (at $x=3$); absolute minimum: $-2$ (at $x=-2$ and $x=1$).
Frequently Asked Questions
How do you construct a sign chart?
Place $x$ on the horizontal axis and divide the domain at points where $f'(x)=0$ or where $f'(x)$ does not exist. Determine the sign of $f'(x)$ on each interval and use arrows to indicate whether $f(x)$ is increasing or decreasing. Include the values of $f(x)$ at boundary points when needed.
How does the second-derivative test work?
If $f'(a)=0$, then $f''(a)>0$ indicates a local minimum, while $f''(a)<0$ indicates a local maximum. If $f''(a)=0$, this test is inconclusive. Higher-order derivatives may sometimes determine the result, but they do not always do so.
Can a point where $f'(x)$ is undefined be a candidate for a local extremum?
Yes. Within the domain of a function, points where $f'(x)=0$ or where $f'(x)$ does not exist are candidates for local extrema. For example, $f(x)=|x|$ is not differentiable at $x=0$, but it has a local minimum there. However, such points are not necessarily local extrema.
Exercises
$f(x) = x^4 - 4x^3$ Construct a sign chart and find its local extrema.
$f(x) = xe^{-x}$ Find its local extrema.
Show Solutions
Solution to Exercise 1
$f'(x) = 4x^3 - 12x^2 = 4x^2(x-3)$
The solutions of $f'(x)=0$ are $x=0,3$. We have $f'(x)<0$ for $x<0$ and $0 \lt x \lt 3$, and $f'(x)>0$ for $x>3$.
At $x=0$, the derivative does not change sign, so there is no local extremum ($f(0)=0$).
At $x=3$, the derivative changes from negative to positive, so the function has a local minimum there, with $f(3)=81-108=-27$.
Solution to Exercise 2
$f'(x) = e^{-x} - xe^{-x} = (1-x)e^{-x}$
Solving $f'(x) = 0$ gives $x = 1$.
$f' > 0$ for $x < 1$ and $f' < 0$ for $x > 1$.
The function has a local maximum at $x=1$, with $f(1)=e^{-1}=\dfrac{1}{e}$.