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First Derivative Test allows us to analyze the rate of change of a function and determine whether it is increasing or decreasing within specific intervals. In the domain of calculus, the behavior of functions is a central focus. One essential tool for understanding the characteristics of a function and identifying critical points is the first derivative test.
- The First Derivative Test relies on the concept of differentiation, which enables us to calculate the derivative of a function.
- The derivative represents the instantaneous rate of change of a function at any given point, providing valuable insights into its behavior.
- By examining the sign of the derivative and the behavior of the function in its vicinity, we can get crucial information about local extrema and points of inflection.
Read More: Applications of Derivatives
| Table of Content |
Key Terms: Function, Derivative, Local Maxima, Local Minima, Point of Inflection, Critical Point, Sign Change, Intervals.
What is First Derivative Test?
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The main objective of the First Derivative Test is to classify critical points as either relative maximums, relative minimums, or neither. A critical point occurs where the derivative of a function equals zero or is undefined. These points serve as potential turning points for the function and can significantly impact its overall shape and behavior.
- To apply the First Derivative Test, first the critical points of the function needs to be found by setting its derivative equal to zero or identifying points where the derivative does not exist.
- Then, then the sign changes in the derivative around these critical points need to be analyzed. If the derivative changes from positive to negative, the function is decreasing and may have a relative maximum at that point.
- Conversely, if the derivative changes from negative to positive, the function is increasing and may have a relative minimum.
- Additionally, the First Derivative Test helps identify points of inflection, which are locations where the concavity of a function changes.
- By examining the behavior of the derivative, the determination of whether the function changes concavity from concave up to concave down or vice versa is possible.
Read More: Continuity and Differentiability
First Derivative Test for Maxima and Minima
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Local Maximum: A point on a function is considered a local maximum if the function reaches its highest value at that point within a small interval around it.
- In other words, a function has a local maximum at a specific point if the values of the function are greater than or equal to the values of the function at nearby points.
Local Minimum: A point on a function is considered a local minimum if the function reaches its lowest value at that point within a small interval around it.
- In other words, a function has a local minimum at a specific point if the values of the function are smaller than or equal to the values of the function at nearby points.
Point of Inflection: A point on a function is called a point of inflection if the concavity of the function changes at that point.
- In other words, the function transitions from being concave up to concave down, or vice versa, at a point of inflection. At this point, the derivative of the function may exist, but it does not necessarily equal zero.
The First Derivative Test is used to identify whether a critical point of a function corresponds to a local maximum or local minimum. This test relies on analyzing the sign changes of the first derivative of the function. Here are steps to follow to identify local maxima and local minima:
- Find the critical points: Determine the values of x where the derivative of the function is equal to zero or undefined.
- These critical points are potential candidates for local extrema or points of inflection.
Evaluate the sign changes: Determine the sign of the derivative on either side of each critical point.
- This can be done by selecting test points within intervals around the critical point and evaluating the sign of the derivative at those points.
Classify the critical points:
- If the derivative changes from positive to negative as we move from left to right around a critical point, the function has a local maximum at that point.
- If the derivative changes from negative to positive as we move from left to right around a critical point, the function has a local minimum at that point.
- If the derivative does not change sign (remains positive or remains negative) or if the derivative is undefined at the critical point, the First Derivative Test is inconclusive, and the critical point may not correspond to a local maximum or local minimum.
It is essential to note that the First Derivative Test only applies to isolated critical points where the derivative is zero or undefined. It does not provide information about global extrema or points of inflection.
- Points of inflection are characterized by changes in the concavity of the function.
- To identify points of inflection, it is necessary to analyze the behavior of the second derivative of the function.
- If the second derivative changes sign at a particular point, that point is a potential point of inflection.
- However, the First Derivative Test does not directly determine points of inflection.
Read More: Approximations
Steps for First Derivative Test
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- Find the derivative: Calculate the derivative of the given function, denoted as f'(x). This step involves differentiating the function f(x) using differentiation rules and techniques.
- Set f'(x) equal to zero: Find the values of x where the derivative f'(x) is equal to zero. Solve the equation f'(x) = 0 to determine these critical points.
- Determine the critical points: Solve the equation obtained in step 2 to find the x-values corresponding to the critical points. These are the points where the derivative is zero.
- Analyze the sign of f'(x): Select test points within intervals on either side of each critical point. Evaluate the sign of f'(x) at these test points.
- Classify the critical points: Based on the sign changes of f'(x), classify each critical point as follows:
- If f'(x) changes from positive to negative, the function f(x) has a local maximum at that critical point.
- If f'(x) changes from negative to positive, the function f(x) has a local minimum at that critical point.
- If there is no sign change, the First Derivative Test is inconclusive, and the critical point may not correspond to a local maximum or local minimum.
- Identify points of inflection (if applicable): If the First Derivative Test is inconclusive at a critical point, it may be a point of inflection. To determine points of inflection, analyze the behavior of the function f(x) in the vicinity of the critical point, considering the concavity changes and the behavior of the second derivative f''(x).
Read More: Increasing and decreasing functions
Conditions for Maxima and Minima
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The conditions for minima and maxima in calculus are determined by the behavior of the derivative of a function. Specifically, there are two conditions that need to be satisfied at a critical point for it to correspond to a local minimum or maximum:
First Derivative Test: At a critical point where the derivative of a function is zero or undefined, the First Derivative Test examines the sign changes of the derivative in the neighborhood of the critical point. The conditions for minima and maxima are as follows:
- Local Minimum: If the derivative changes from positive to negative as we move from left to right around a critical point, the function has a local minimum at that point. This implies that the function is decreasing before the critical point and increasing after it.
- Local Maximum: If the derivative changes from negative to positive as we move from left to right around a critical point, the function has a local maximum at that point. This implies that the function is increasing before the critical point and decreasing after it.
Second Derivative Test: In addition to the First Derivative Test, the Second Derivative Test analyzes the concavity of a function near a critical point. The second derivative of a function provides information about its concavity. The conditions for minima and maxima based on the second derivative are as follows:
- Local Minimum: If the second derivative of a function is positive at a critical point, the function has a local minimum at that point. This indicates that the function is concave up around the critical point.
- Local Maximum: If the second derivative of a function is negative at a critical point, the function has a local maximum at that point. This indicates that the function is concave down around the critical point.
Note: If the second derivative is zero or undefined at a critical point, the Second Derivative Test is inconclusive, and further analysis is needed.
Read More: Tangents and normals
Applications of First Derivative Test
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The First Derivative Test is a valuable tool in calculus with various applications in mathematical analysis and optimization. Some of the important uses and applications of the First Derivative Test include:
Identifying local extrema:
- The First Derivative Test helps determine whether a critical point corresponds to a local minimum or local maximum of a function.
- By analyzing the sign changes of the derivative around critical points, the presence and location of these extrema can be identified.
Optimization problems:
- The First Derivative Test is commonly used to solve optimization problems.
- By finding critical points and applying the First Derivative Test, the optimal values or conditions that maximize or minimize a given objective function can be determined.
- This has applications in areas such as economics, physics, engineering, and management science.
Curve sketching:
- The First Derivative Test provides valuable information about the behavior of a function, helping in the process of sketching its graph.
- By analyzing the sign changes of the derivative and identifying local extrema, accurately the shape and characteristics of the function can be depicted.
Determining increasing and decreasing intervals:
- By analyzing the sign of the derivative, the First Derivative Test allows us to determine the intervals on which a function is increasing or decreasing.
- This information is useful for understanding the overall trend and behavior of a function.
Concavity and points of inflection:
- Although the First Derivative Test primarily focuses on identifying extrema, it can also be used to provide insights into the concavity of a function and locate points of inflection.
- By combining the analysis of the first and second derivatives, the concave-up and concave-down regions of a function can be determined, and points where the concavity changes can be located.
Root-finding and solving equations:
- The First Derivative Test can be applied in solving equations by finding the critical points where the derivative is zero.
- This information can be useful in finding the roots of equations or solving problems that involve finding specific values of x.
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Solved Example
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Example: Find the critical points, analyze the sign changes of the derivative, and classify the critical points using the First Derivative Test for the function: f(x) = x3 - 4x2 + 3x + 2
Solution: Step 1: Find the derivative:
f'(x) = 3x2 - 8x + 3
Step 2: Set f'(x) equal to zero and solve for critical points:
3x2 - 8x + 3 = 0
Factoring or applying the quadratic formula,
(x - 1)(3x - 3) = 0
This gives two critical points:
x = 1 and x = 1/3
Step 3: Determine the critical points:
The critical points are x = 1 and x = 1/3.
Step 4: Analyze the sign of f'(x):
To analyze the sign changes, test points need to be chosed within intervals around the critical points. Let's select x = 0, x = 1/2, and x = 2 as test points.
For x < 1/3:
f'(x) = 3x2 - 8x + 3
f'(-1) = 3(-1)2 - 8(-1) + 3 = 14 (positive)
f'(1/2) = 3(1/2)2 - 8(1/2) + 3 = -1.25 (negative)
For 1/3 < x < 1:
f'(1/2) = -1.25 (negative)
f'(2) = 3(2)2 - 8(2) + 3 = -5 (negative)
For x > 1:
f'(2) = -5 (negative)
f'(3) = 3(3)2 - 8(3) + 3 = 10 (positive)
Step 5: Classify the critical points:
At x = 1, f'(x) changes from negative to positive, indicating a local minimum at x = 1.
At x = 1/3, f'(x) does not change sign, so the First Derivative Test is inconclusive at this point.
Step 6: Identify points of inflection (if applicable):
Since the First Derivative Test is inconclusive at x = 1/3, analyzation of the behavior of the second derivative f''(x) needs to be done to determine if x = 1/3 is a point of inflection.
Taking the derivative of f'(x), f''(x) = 6x - 8
Evaluating f''(1/3), we get:
f''(1/3) = 6(1/3) - 8 = -6 - 8 = -14 (negative)
Since f''(1/3) is negative, x = 1/3 is a point of inflection.
Therefore, for the given function f(x) = x3 - 4x2 + 3x + 2:
x = 1 is a local minimum.
x = 1/3 is a point of inflection.
Read More: Differential Equations
Things to Remember
- Critical points are the values of x where the derivative of the function is zero or undefined. These points are potential candidates for local extrema or points of inflection.
- The First Derivative Test relies on analyzing the sign changes of the derivative around critical points.
- If the derivative changes sign from positive to negative as we move from left to right around a critical point, the function has a local minimum at that point.
- If the derivative changes sign from negative to positive as we move from left to right around a critical point, the function has a local maximum at that point.
- If the derivative does not change sign at a critical point, the First Derivative Test is inconclusive. In this case, the critical point may not correspond to a local extrema.
- To apply the First Derivative Test, select test points within intervals on either side of each critical point.
Sample Questions
Ques: Determine the intervals on which the function f(x) = x3 - 3x2 + 2x is increasing or decreasing. Identify any local extrema. (3 Marks)
Ans: Step 1: Find the derivative: f'(x) = 3x2 - 6x + 2.
Step 2: Set f'(x) = 0 and solve for x: 3x2 - 6x + 2 = 0. The solutions are x = 1 ± (2)/3.
Step 3: Choose test points within the intervals and analyze the sign of f'(x).
For x < 1 - (2)/3, f'(x) is positive, indicating an increasing interval.
For 1 - (2)/3 < x < 1 + (2)/3, f'(x) is negative, indicating a decreasing interval.
For x > 1 + (2)/3, f'(x) is positive again, indicating an increasing interval.
There are no local extrema.
Ques: Find the local extrema of the function f(x) = 2x3 - 9x2 + 12x - 1. (2 Marks)
Ans: Step 1: Find the derivative: f'(x) = 6x2 - 18x + 12.
Step 2: Set f'(x) = 0 and solve for x: 6x2 - 18x + 12 = 0. Simplifying gives x2 - 3x + 2 = 0. Factoring yields (x - 1)(x - 2) = 0. The critical points are x = 1 and x = 2.
Step 3: Choose test points within the intervals and analyze the sign of f'(x).
For x < 1, f'(x) is positive, indicating an increasing interval.
For 1 < x < 2, f'(x) is negative, indicating a decreasing interval.
For x > 2, f'(x) is positive, indicating an increasing interval.
Therefore, the function has a local minimum at x = 1.
Ques: Determine the intervals on which the function f(x) = x4 - 4x3 - 6x2 + 24x is increasing or decreasing. Identify any local extrema. (3 Marks)
Ans: Step 1: Find the derivative: f'(x) = 4x3 - 12x2 - 12x + 24.
Step 2: Set f'(x) = 0 and solve for x. By factoring, x = -2, x = 1, and x = 2 are the critical points.
Step 3: Choose test points within the intervals and analyze the sign of f'(x).
For x < -2, f'(x) is positive, indicating an increasing interval.
For -2 < x < 1, f'(x) is negative, indicating a decreasing interval.
For 1 < x < 2, f'(x) is positive, indicating an increasing interval.
For x > 2, f'(x) is negative, indicating a decreasing interval.
Thus, the function has a local maximum at x = -2 and a local minimum at x = 2.
Ques: Find the local extrema of the function f(x) = x5 - 5x4 + 10x3 - 10x + 2. (2 Marks)
Ans: Step 1: Find the derivative: f'(x) = 5x4 - 20x3 + 30x2 - 10.
Step 2: Set f'(x) = 0 and solve for x. By factoring, we fin x = 1 is the critical point.
Step 3: Choose test points within the intervals and analyze the sign of f'(x).
For x < 1, f'(x) is positive, indicating an increasing interval.
For x > 1, f'(x) is negative, indicating a decreasing interval.
Hence, the function has a local maximum at x = 1.
Ques: Determine the intervals on which the function f(x) = x3 - 3x2 - 9x + 5 is increasing or decreasing. Identify any local extrema. (2 Marks)
Ans: Step 1: Find the derivative: f'(x) = 3x2 - 6x - 9.
Step 2: Set f'(x) = 0 and solve for x. By factoring, x = -1 and x = 3 are the critical points.
Step 3: Choose test points within the intervals and analyze the sign of f'(x).
For x < -1, f'(x) is positive, indicating an increasing interval.
For -1 < x < 3, f'(x) is negative, indicating a decreasing interval.
For x > 3, f'(x) is positive, indicating an increasing interval.
Therefore, the function has a local minimum at x = -1.
Ques: Find the local extrema of the function f(x) = x3 + 6x2 - 9x - 10. (2 Marks)
Ans: Step 1: Find the derivative: f'(x) = 3x2 + 12x - 9.
Step 2: Set f'(x) = 0 and solve for x. By factoring, x = -3 and x = 1 are the critical points.
Step 3: Choose test points within the intervals and analyze the sign of f'(x).
For x < -3, f'(x) is positive, indicating an increasing interval.
For -3 < x < 1, f'(x) is negative, indicating a decreasing interval.
For x > 1, f'(x) is positive, indicating an increasing interval.
Hence, the function has a local minimum at x = 1 and a local maximum at x = -3.
Ques: Determine the intervals on which the function f(x) = x4 - 8x2 + 12 is increasing or decreasing. Identify any local extrema. (2 Marks)
Ans: Step 1: Find the derivative: f'(x) = 4x3 - 16x.
Step 2: Set f'(x) = 0 and solve for x. By factoring, x = 0 and x = ±2 are the critical points.
Step 3: Choose test points within the intervals and analyze the sign of f'(x).
For x < -2, f'(x) is positive, indicating an increasing interval.
For -2 < x < 0, f'(x) is negative, indicating a decreasing interval.
For 0 < x < 2, f'(x) is positive, indicating an increasing interval.
For x > 2, f'(x) is positive, indicating an increasing interval.
Thus, the function has a local minimum at x = -2 and a local maximum at x = 2.
Ques: Find the local extrema of the function f(x) = 3x3 - 12x2 - 15x + 10. (2 Marks)
Ans: Step 1: Find the derivative: f'(x) = 9x2 - 24x - 15.
Step 2: Set f'(x) = 0 and solve for x. By factoring, x = -1 and x = 5/3 are the critical points.
Step 3: Choose test points within the intervals and analyze the sign of f'(x).
For x < -1, f'(x) is positive, indicating an increasing interval.
For -1 < x < 5/3, f'(x) is negative, indicating a decreasing interval.
For x > 5/3, f'(x) is positive, indicating an increasing interval.
Hence, the function has a local minimum at x = -1.
Ques: Determine the intervals on which the function f(x) = x3 - 12x2 + 36x - 24 is increasing or decreasing. Identify any local extrema. (2 Marks)
Ans: Step 1: Find the derivative: f'(x) = 3x2 - 24x + 36.
Step 2: Set f'(x) = 0 and solve for x. By factoring, x = 2 and x = 6 are the critical points.
Step 3: Choose test points within the intervals and analyze the sign of f'(x).
For x < 2, f'(x) is negative, indicating a decreasing interval.
For 2 < x < 6, f'(x) is positive, indicating an increasing interval.
For x > 6, f'(x) is negative, indicating a decreasing interval.
Therefore, the function has a local maximum at x = 2 and a local minimum at x = 6.
Ques: Find the local extrema of the function f(x) = x4 - 4x3 + 4x2 + 2. (2 Marks)
Ans: Step 1: Find the derivative: f'(x) = 4x3 - 12x2 + 8x.
Step 2: Set f'(x) = 0 and solve for x. By factoring, x = 0, x = 1, and x = 2 are the critical points.
Step 3: Choose test points within the intervals and analyze the sign of f'(x).
For x < 0, f'(x) is negative, indicating a decreasing interval.
For 0 < x < 1, f'(x) is positive, indicating an increasing interval.
For 1 < x < 2, f'(x) is negative, indicating a decreasing interval.
For x > 2, f'(x) is positive, indicating an increasing interval.
Thus, the function has a local minimum at x = 0 and a local maximum at x = 2.
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