Lesson 17: Increasing and Decreasing Functions; First Derivative Test
Last time: Critical numbers provide us with candidates for relative extrema. In this lesson, we use the sign of the first derivative to determine where a function is increasing or decreasing, and then apply the first derivative test to classify relative extrema.
17.1 Increasing and Decreasing Functions
Definition: A function \(f\) is increasing on an interval if \(x < y\) implies \(f(x) < f(y)\). That is, as the input increases, the output also increases.
Definition: A function \(f\) is decreasing on an interval if \(x < y\) implies \(f(x) > f(y)\). That is, as the input increases, the output decreases.
Diagram Description: Two side-by-side curve sketches illustrate the definitions. On the left, an increasing curve rises from lower-left to upper-right; tangent lines drawn at several points along the curve all have positive slope. On the right, a decreasing curve falls from upper-left to lower-right; tangent lines at several points all have negative slope. This reinforces the link between the sign of the derivative and whether the function increases or decreases.
Theorem: Let \(f(x)\) be a continuous and differentiable function on an open interval \(I\).
- If \(f'(x) > 0\) for all \(x\) in \(I\), then \(f(x)\) is increasing on \(I\).
- If \(f'(x) < 0\) for all \(x\) in \(I\), then \(f(x)\) is decreasing on \(I\).
General Strategy for Finding Intervals of Increase/Decrease:
- Compute \(f'(x)\).
- Find all critical numbers (where \(f'(x) = 0\) or \(f'(x)\) is undefined).
- Use the critical numbers to divide the real line into open intervals.
- Pick a test value in each interval and evaluate \(f'\) at that value.
- If \(f' > 0\) on the interval, \(f\) is increasing; if \(f' < 0\), \(f\) is decreasing.
Example 1: Finding Intervals of Increase and Decrease
Suppose \(f(x) = x^3 - 9x^2 - 21x\). Find the open intervals on which \(f(x)\) is increasing or decreasing.
Solution:
First, we compute the derivative of \(f(x)\):
\[f'(x) = 3x^2 - 18x - 21 = 3(x^2 - 6x - 7)\]Next, we set \(f'(x) = 0\) to find the critical numbers:
\[3(x^2 - 6x - 7) = 0 \implies 3(x-7)(x+1) = 0 \implies x = 7, \; x = -1\]These two critical numbers divide the real line into three intervals: \((-\infty, -1)\), \((-1, 7)\), and \((7, \infty)\). We test a \(x\)-value from each interval by plugging the \(x\)-value into \(f'(x)\).
| Interval | Test Value | \(f'(\text{test value})\) | Sign | Conclusion |
|---|---|---|---|---|
| \((-\infty, -1)\) | \(x = -2\) | \(3\cdot (4 + 12 - 7) = 3\cdot 9 = 27 > 0\) | \(+\) | Increasing |
| \((-1, 7)\) | \(x = 0\) | \(f'(0) = -21 < 0\) | \(-\) | Decreasing |
| \((7, \infty)\) | \(x = 8\) | \(3\cdot (64 - 48 - 7) = 3\cdot 9 = 27 > 0\) | \(+\) | Increasing |
Sign Chart Description: A number line is marked with the critical numbers \(-1\) and \(7\). To the left of \(-1\) the sign of \(f'\) is \(+\) (arrow pointing up-right). Between \(-1\) and \(7\) the sign is \(-\) (arrow pointing down-right). To the right of \(7\) the sign is \(+\) (arrow pointing up-right).
Increasing interval: \((-\infty, -1)\cup (7, \infty)\)
Decreasing interval: \((-1, 7)\)
17.2 The First Derivative Test
Now that we can determine intervals of increase and decrease, we can classify the critical numbers as relative maxima, relative minima, or neither.
First Derivative Test: Let \(c\) be a critical number of a function \(f(x)\) that is continuous on an open interval \(I\) containing \(c\). We examine the sign of \(f'(x)\) on each side of \(x = c\).
- Scenario 1 (Relative Maximum): If \(f'(x) > 0\) to the left of \(c\) and \(f'(x) < 0\) to the right of \(c\), then \(f\) has a relative maximum at \(x = c\).
- Scenario 2 (Relative Minimum): If \(f'(x) < 0\) to the left of \(c\) and \(f'(x) > 0\) to the right of \(c\), then \(f\) has a relative minimum at \(x = c\).
- Scenario 3 (No Extremum): If \(f'(x) > 0\) on both sides of \(c\), then \(f\) has no relative extremum at \(x = c\).
- Scenario 4 (No Extremum): If \(f'(x) < 0\) on both sides of \(c\), then \(f\) has no relative extremum at \(x = c\).
Diagram Description (Four Scenarios): Four pairs of sketches illustrate each scenario. In Scenario 1, the curve rises to a peak at \((c, f(c))\) and then falls — a classic hill shape, labeled "rel. max. at \(x = c\)." In Scenario 2, the curve falls to a valley at \((c, f(c))\) and then rises — a classic valley shape, labeled "rel. min. at \(x = c\)." In Scenarios 3 and 4, the curve passes through \((c, f(c))\) without changing direction (continuing to increase or continuing to decrease, respectively), so no extremum occurs.
Example 1 (Revisited)
Find the relative extrema of \(f(x) = x^3 - 9x^2 - 21x\).
Solution:
From Example 1, the critical numbers are \(x = -1\) and \(x = 7\), and the sign chart shows \(f'\) changes from \(+\) to \(-\) at \(x = -1\), and from \(-\) to \(+\) at \(x = 7\).
Sign Chart Description: The number line from Example 1 is reused. At \(x = -1\), the sign changes from \(+\) (increasing) to \(-\) (decreasing), indicating a relative maximum. At \(x = 7\), the sign changes from \(-\) (decreasing) to \(+\) (increasing), indicating a relative minimum. A small sketch shows a hill at the left critical number and a valley at the right.
By the first derivative test, \(f'\) changes from \(+\) to \(-\) at \(x = -1\), and so we have a relative maximum at \((-1,\; f(-1))\). Additionally, \(f'\) changes from \(-\) to \(+\) at \(x = 7\), and so we have a relative minimum at \((7,\; f(7))\).
Computing the function outputs at these \(x\)-values:
\[f(-1) = (-1)^3 - 9\cdot (-1)^2 - 21\cdot (-1) = -1 - 9 + 21 = 11\] \[f(7) = 7^3 - 9\cdot 7^2 - 21\cdot 7 = 343 - 441 - 147 = -245\]Relative maximum: \((-1,\; 11)\)
Relative minimum: \((7,\; -245)\)
Example 2
Suppose \(f(x) = 10x^2 - 4x\). Find the open intervals on which \(f(x)\) is increasing or decreasing, as well as the relative extrema, if any exist.
Solution:
First, we compute the derivative:
\[f'(x) = 20x - 4\]Next, we set \(f'(x) = 0\):
\[20x - 4 = 0 \implies x = \frac{1}{5}\]So, \(x = \frac{1}{5}\) is the only critical number, dividing the real line into \(\left(-\infty, \frac{1}{5}\right)\) and \(\left(\frac{1}{5}, \infty\right)\). Testing:
| Interval | Test Value | \(f'(\text{test value})\) | Sign | Conclusion |
|---|---|---|---|---|
| \(\left(-\infty, \frac{1}{5}\right)\) | \(x = 0\) | \(f'(0) = -4 < 0\) | \(-\) | Decreasing |
| \(\left(\frac{1}{5}, \infty\right)\) | \(x = 1\) | \(f'(1) = 16 > 0\) | \(+\) | Increasing |
Since \(f'\) changes from \(-\) to \(+\) at \(x = \frac{1}{5}\), the first derivative test implies that we have a relative minimum at \(x=\frac{1}{5}\). Note that:
\[f\!\left(\tfrac{1}{5}\right) = 10\cdot \left(\tfrac{1}{5}\right)^{\!2} - 4\cdot \left(\tfrac{1}{5}\right) = \frac{10}{25} - \frac{4}{5} = \frac{2}{5} - \frac{4}{5} = -\frac{2}{5}\]Increasing interval: \(\left(\frac{1}{5}, \infty\right)\)
Decreasing interval: \(\left(-\infty, \frac{1}{5}\right)\)
Relative minimum: \(\left(\frac{1}{5},\; -\frac{2}{5}\right)\)
Example 3
Suppose \(f(x) = x^3 e^{2x+6}\). Find the open intervals on which \(f(x)\) is increasing or decreasing, as well as the relative extrema, if any exist.
Solution:
Applying the Product Rule, we have that:
\[f'(x) = 3x^2 e^{2x+6} + 2e^{2x+6} x^3 = x^2 e^{2x+6}(3 + 2x)\]Setting \(f'(x) = 0\):
\[x^2 e^{2x+6}(3 + 2x) = 0 \implies x = 0 \quad \text{or} \quad x = -\frac{3}{2}\]These are the critical numbers. They divide the real line into three intervals. Testing:
| Interval | Test Value | \(f'(\text{test value})\) | Sign | Conclusion |
|---|---|---|---|---|
| \(\left(-\infty, -\frac{3}{2}\right)\) | \(x = -2\) | \(4\cdot e^{2}\cdot (-1)<0\) | \(-\) | Decreasing |
| \(\left(-\frac{3}{2}, 0\right)\) | \(x = -1\) | \(1\cdot e^4\cdot 1>0\) | \(+\) | Increasing |
| \((0, \infty)\) | \(x = 1\) | \(1\cdot e^8\cdot 5>0\) | \(+\) | Increasing |
At \(x = -\frac{3}{2}\), \(f'\) changes from \(-\) to \(+\), and so we have a relative minimum. At \(x = 0\), \(f'\) does not change sign (positive on both sides), and so we have no extremum. We see that:
\[f\!\left(-\tfrac{3}{2}\right) = \left(-\tfrac{3}{2}\right)^{\!3} \cdot e^{2(-3/2)+6} = -\frac{27}{8}\, e^{3}\]Increasing interval: \(\left(-\frac{3}{2}, \infty\right)\)
Decreasing interval: \(\left(-\infty, -\frac{3}{2}\right)\)
Relative minimum: \(\left(-\dfrac{3}{2},\; -\dfrac{27}{8}e^3\right)\)
Example 4
Suppose \(f'(x) = e^{11x}(2x^2 - 26)\). Find the open intervals on which \(f(x)\) is increasing or decreasing, as well as the \(x\)-values of the relative extrema, if any exist.
Solution:
We set \(f'(x) = 0\):
\[e^{11x}(2x^2 - 26)=0\implies 2x^2 - 26 = 0 \implies x^2 = 13 \implies x = \pm\sqrt{13}\]These critical numbers divide the real line into three intervals. Testing:
| Interval | Test Value | \(f'(\text{test value})\) | Sign | Conclusion |
|---|---|---|---|---|
| \((-\infty, -\sqrt{13})\) | \(x = -4\) | \(2\cdot 16-26 = 6 > 0\) | \(+\) | Increasing |
| \((-\sqrt{13}, \sqrt{13})\) | \(x = 0\) | \(2\cdot 0-26 = -26 < 0\) | \(-\) | Decreasing |
| \((\sqrt{13}, \infty)\) | \(x = 4\) | \(2\cdot 16-26 = 6 > 0\) | \(+\) | Increasing |
At \(x = -\sqrt{13}\), \(f'\) changes from \(+\) to \(-\), and so we have a relative maximum at \(x = -\sqrt{13}\). At \(x = \sqrt{13}\), \(f'\) changes from \(-\) to \(+\), and so we have a relative minimum at \(x = \sqrt{13}\).
Increasing interval: \(\left(-\infty, -\sqrt{13}\right) \cup \left(\sqrt{13}, \infty\right)\)
Decreasing interval: \(\left(-\sqrt{13}, \sqrt{13}\right)\)
Relative maximum at \(x = -\sqrt{13}\)
Relative minimum at \(x = \sqrt{13}\)
Example 5
The critical numbers of \(f(x) = 5\sin(x)\) on the interval \((0, 2\pi)\) are \(x = \dfrac{\pi}{2}\) and \(x = \dfrac{3\pi}{2}\). What are the \(x\)-values on \((0, 2\pi)\) at which \(f(x)\) has a relative maximum?
Solution:
Compute the derivative:
\[f'(x) = 5\cos(x)\]The critical numbers \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\) divide \((0, 2\pi)\) into three intervals. Test a value from each:
| Interval | Test Value | \(f'(\text{test value})\) | Sign | Conclusion |
|---|---|---|---|---|
| \(\left(0, \frac{\pi}{2}\right)\) | \(x = \frac{\pi}{4}\) | \(5\cos\!\left(\frac{\pi}{4}\right) = \frac{5\sqrt{2}}{2} > 0\) | \(+\) | Increasing |
| \(\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)\) | \(x = \pi\) | \(5\cos(\pi) = -5 < 0\) | \(-\) | Decreasing |
| \(\left(\frac{3\pi}{2}, 2\pi\right)\) | \(x = \frac{7\pi}{4}\) | \(5\cos\!\left(\frac{7\pi}{4}\right) = \frac{5\sqrt{2}}{2} > 0\) | \(+\) | Increasing |
Sign Chart Description: A number line over the interval \((0, 2\pi)\) is marked at \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\). The sign of \(f'\) is \(+\) on \(\left(0, \frac{\pi}{2}\right)\), then \(-\) on \(\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)\), then \(+\) again on \(\left(\frac{3\pi}{2}, 2\pi\right)\). Arrows pointing up-right and down-right indicate direction of the function.
At \(x = \frac{\pi}{2}\), \(f'\) changes from \(+\) to \(-\), and so we have a relative maximum. At \(x = \frac{3\pi}{2}\), \(f'\) changes from \(-\) to \(+\), and so we have a relative minimum.
Relative maximum at \(x = \dfrac{\pi}{2}\)
Key Takeaways: The sign of \(f'(x)\) completely determines where \(f\) is increasing or decreasing. Critical numbers are where the sign of \(f'\) can change -- checking the sign on each side via a sign chart lets us classify each critical number as a relative maximum, relative minimum, or neither using the first derivative test. Remember that a critical number can occur without producing an extremum if the sign of \(f'\) does not change at that critical number.