3.3: The First Derivative Test
Learning Objectives
A student will be able to:
- Find intervals where a function is increasing and decreasing.
- Apply the First Derivative Test to find extrema and sketch graphs.
Introduction
In this lesson we will discuss increasing and decreasing properties of functions, and introduce a method with which to study these phenomena, the First Derivative Test. This method will enable us to identify precisely the intervals where a function is either increasing or decreasing, and also help us to sketch the graph. Note on notation: The symbol \begin{align*}\epsilon\end{align*} and \begin{align*} \in \end{align*} are equivalent and denote that a particular element is contained within a particular set.
- Definition
- A function \begin{align*}f\end{align*} is said to be increasing on \begin{align*}[a,b]\end{align*} contained in the domain of \begin{align*}f\end{align*} if \begin{align*}f(x_1) \le f(x_2)\end{align*} whenever \begin{align*}x_{1} \le x_{2}\end{align*} for all \begin{align*}x_1, x_2 \in [a,b].\end{align*} A function \begin{align*}f\end{align*} is said to be decreasing on \begin{align*}[a,b]\end{align*} contained in the domain of \begin{align*}f\end{align*} if \begin{align*}f(x_1) \ge f(x_2)\end{align*} whenever \begin{align*}x_1 \ge x_2\end{align*} for all \begin{align*}x_1, x_2 \in [a,b].\end{align*}
- If \begin{align*}f(x_1) < f(x_2)\end{align*} whenever \begin{align*}x_1 < x_2\end{align*} for all \begin{align*}x_1, x_2 \in [a,b],\end{align*} then we say that \begin{align*}f\end{align*} is strictly increasing on \begin{align*}[a,b].\end{align*} If \begin{align*}f(x_1) > f(x_2)\end{align*} whenever \begin{align*}x_1 > x_2\end{align*} for all \begin{align*}x_1, x_2 \in [a,b],\end{align*} then we say that \begin{align*}f\end{align*} is strictly decreasing on \begin{align*}[a,b].\end{align*}
- We saw several examples in the Lesson on Extreme and the Mean Value Theorem of functions that had these properties.
Example 1:
The function \begin{align*}f(x) = x^3\end{align*} is strictly increasing on \begin{align*}(-\infty, +\infty)\end{align*}:
Example 2:
The function indicated here is strictly increasing on \begin{align*}(0,a)\end{align*} and \begin{align*}(b,c)\end{align*} and strictly decreasing on \begin{align*}(a,b)\end{align*} and \begin{align*}(c,d).\end{align*}
We can now state the theorems that relate derivatives of functions to the increasing/decreasing properties of functions.
Theorem: If \begin{align*}f\end{align*} is continuous on interval \begin{align*}[a,b],\end{align*} then:
- If \begin{align*}f'(x) > 0\end{align*} for every \begin{align*}x \in [a,b],\end{align*} then \begin{align*}f\end{align*} is strictly increasing in \begin{align*}[a,b].\end{align*}
- If \begin{align*}f'(x) < 0\end{align*} for every \begin{align*}x \in [a,b],\end{align*} then \begin{align*}f\end{align*} is strictly decreasing in \begin{align*}[a,b].\end{align*}
Proof: We will prove the first statement. A similar method can be used to prove the second statement and is left as an exercise to the student.
Consider \begin{align*}x_1, x_2 \in [a,b]\end{align*} with \begin{align*}x_1 < x_2.\end{align*} By the Mean Value Theorem, there exists \begin{align*}c \in (x_1,x_2)\end{align*} such that
\begin{align*}f(x_2) - f(x_1) = (x_2 - x_1) f'(c).\end{align*}
By assumption, \begin{align*}f'(x) > 0\end{align*} for every \begin{align*}x \in [a,b]\end{align*}; hence \begin{align*}f'(c) > 0.\end{align*} Also, note that \begin{align*}x_2 - x_1 > 0.\end{align*}
Hence \begin{align*}f(x_2) - f(x_1) > 0\end{align*} and \begin{align*}f(x_2) > f(x_1).\end{align*}
We can observe the consequences of this theorem by observing the tangent lines of the following graph. Note the tangent lines to the graph, one in each of the intervals \begin{align*}(0,a), (a,b),\end{align*} \begin{align*}(b,+\infty).\end{align*}
Note first that we have a relative maximum at \begin{align*}x = a\end{align*} and a relative minimum at \begin{align*}x = b.\end{align*} The slopes of the tangent lines change from positive for \begin{align*}x \in (0,a)\end{align*} to negative for \begin{align*}x \in (a,b)\end{align*} and then back to positive for \begin{align*}x \in (b, +\infty)\end{align*}. From this we example infer the following theorem:
First Derivative Test
Suppose that \begin{align*}f\end{align*} is a continuous function and that \begin{align*}x = c\end{align*} is a critical value of \begin{align*}f.\end{align*} Then:
- If \begin{align*}f'\end{align*} changes from positive to negative at \begin{align*}x = c,\end{align*} then \begin{align*}f\end{align*} has a local maximum at \begin{align*}x = c.\end{align*}
- If \begin{align*}f'\end{align*} changes from negative to positive at \begin{align*}x = c,\end{align*} then \begin{align*}f\end{align*} has a local minimum at \begin{align*}x = c.\end{align*}
- If \begin{align*}f'\end{align*} does not change sign at \begin{align*}x = c,\end{align*} then \begin{align*}f\end{align*} has neither a local maximum nor minimum at \begin{align*}x = c.\end{align*}
Proof of these three conclusions is left to the reader.
Example 3:
Our previous example showed a graph that had both a local maximum and minimum. Let’s reconsider \begin{align*}f(x) = x^3\end{align*} and observe the graph around \begin{align*}x = 0.\end{align*} What happens to the first derivative near this value?
We observe that the tangent lines to the graph are positive on both sides of \begin{align*}x=0\end{align*}. The first derivative test (\begin{align*}f'(x)=3x^2\end{align*}) verifies this fact, and that the slopes of the tangent line are positive for all nonzero \begin{align*}x\end{align*}. Although \begin{align*}f'(0)=0\end{align*}, and so \begin{align*}f\end{align*} has a critical value at \begin{align*}x=0\end{align*}, the third part of the First Derivative Test tells us that the failure of \begin{align*}f'\end{align*} to change sign at \begin{align*}x=0\end{align*} means that \begin{align*}f\end{align*} has neither a local minimum nor a local maximum at \begin{align*}x=0\end{align*}.
Example 4:
Let's consider the function \begin{align*}f (x) = x^2 + 6 x - 9\end{align*} and observe the graph around \begin{align*}x = -3.\end{align*} What happens to the first derivative near this value?
We observe that the slopes of the tangent lines to the graph change from negative to positive at \begin{align*}x = -3.\end{align*} The first derivative test verifies this fact. Note that the slopes of the tangent lines to the graph are negative for \begin{align*}x \in (-\infty, -3)\end{align*} and positive for \begin{align*}x \in (-3, -\infty).\end{align*}
Lesson Summary
- We found intervals where a function is increasing and decreasing.
- We applied the First Derivative Test to find extrema and sketch graphs.
Multimedia Links
For more examples on determining whether a function is increasing or decreasing (9.0), see Math Video Tutorials by James Sousa, Determining where a function is increasing and decreasing using the first derivative (10:05).
For a video presentation of increasing and decreasing trigonometric functions and relative extrema (9.0), see Math Video Tutorials by James Sousa, Increasing and decreasing trig functions, relative extrema (6:03).
For more information on finding relative extrema using the first derivative (9.0), see Math Video Tutorials by James Sousa, Finding relative extrema using the first derivative (6:19).
Review Questions
In problems #1–2, identify the intervals where the function is increasing, decreasing, or is constant. (Units on the axes indicate single units).
- Give the sign of the following quantities for the graph in #2.
- \begin{align*}f'(-3)\end{align*}
- \begin{align*}f'(1)\end{align*}
- \begin{align*}f'(3)\end{align*}
- \begin{align*}f'(4)\end{align*}
For problems #4–6, determine the intervals in which the function is increasing and those in which it is decreasing. Sketch the graph.
- \begin{align*}f(x) = x^2 - \frac{1} {x}\end{align*}
- \begin{align*}f(x) = (x^2 - 1)^5\end{align*}
- \begin{align*}f(x) = (x^2 - 1)^4\end{align*}
For problems #7–10:
a. Use the First Derivative Test to find the intervals where the function increases and/or decreases
b. Identify all max, mins, or relative max and mins
c. Sketch the graph
- \begin{align*}f(x) = -x^2 -4x - 1\end{align*}
- \begin{align*}f(x) = x^3 + 3x^2 - 9x + 1\end{align*}
- \begin{align*} f(x) = x^\frac{2} {3} (x - 5)\end{align*}
- \begin{align*} f(x) = 2x{\sqrt{x^2 + 1}}\end{align*}
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