The First Derivative Test: Videos & Practice Problems

To determine where a function is increasing or decreasing, analyze its derivative. A positive derivative indicates an increasing function, while a negative derivative indicates a decreasing function. Critical points, where the derivative equals zero or does not exist, help identify intervals of increase or decrease. The first derivative test reveals local maxima and minima by observing sign changes around critical points. For example, if the derivative changes from positive to negative, a local maximum occurs; if it changes from negative to positive, a local minimum occurs. Understanding these concepts is essential for analyzing function behavior.

Determining Where a Function is Increasing & Decreasing

Understanding the behavior of a function, specifically where it is increasing or decreasing, can be effectively determined using its derivative. When analyzing a function from left to right, it is clear that a function is increasing if it rises and decreasing if it falls. The key to this analysis lies in the derivative, which represents the slope of the tangent line to the function at any given point. A positive derivative indicates that the function is increasing, while a negative derivative signifies that the function is decreasing.

To illustrate this concept, consider the function \( f(x) = -x^2 + 4x + 5 \). The first step is to find the derivative, which can be calculated using the power rule. The derivative is given by:

\[ f'(x) = -2x + 4 \]

Next, to determine whether the function is increasing or decreasing at specific points, such as \( x = 0 \) and \( x = 5 \), we substitute these values into the derivative:

For \( x = 0 \):

\[ f'(0) = -2(0) + 4 = 4 \] (positive)

This indicates that the function is increasing at \( x = 0 \).

For \( x = 5 \):

\[ f'(5) = -2(5) + 4 = -6 \] (negative)

This indicates that the function is decreasing at \( x = 5 \).

To find the intervals where the function is increasing or decreasing, we first identify the critical points, which occur where the derivative is zero or undefined. Setting the derivative equal to zero:

\[ -2x + 4 = 0 \]

Solving for \( x \) gives:

\[ -2x = -4 \Rightarrow x = 2 \]

Now, we create a sign chart to analyze the intervals defined by the critical point \( x = 2 \). This divides the number line into two intervals: \( (-\infty, 2) \) and \( (2, \infty) \). We can choose test values from each interval to determine the sign of the derivative:

For the interval \( (-\infty, 2) \), we can choose \( x = 0 \):

\[ f'(0) = 4 \] (positive)

This indicates that the function is increasing on the interval \( (-\infty, 2) \).

For the interval \( (2, \infty) \), we can choose \( x = 5 \):

\[ f'(5) = -6 \] (negative)

This indicates that the function is decreasing on the interval \( (2, \infty) \).

In summary, the function \( f(x) = -x^2 + 4x + 5 \) is increasing on the interval \( (-\infty, 2) \) and decreasing on the interval \( (2, \infty) \). By following these steps—finding the derivative, identifying critical points, creating a sign chart, and testing intervals—you can effectively determine where a function is increasing or decreasing.

Identify the open intervals on which the function is increasing or decreasing.
$f(x)=3x^4+8x^3-18x^2+7$

• A. Increasing on $\left(-\infty,-3\right)\&\left(0,1\right)$, Decreasing on $\left(-3,0\right)\&\left(1,\infty\right)$
• B. Increasing on $\left(-3,0\right)\&\left(1,\infty\right)$, Decreasing on $\left(-\infty,-3\right)\&\left(0,1\right)$
• C. Increasing on $\left(-\infty,0\right)$, Decreasing on $\left(0,\infty\right)$
• D. Increasing on $\left(0,\infty\right)$, Decreasing on $\left(-\infty,0\right)$

Answer: (B)

Identify the open intervals on which the function is increasing or decreasing.
$f(x)=x^{2/3}(4-x)$

• A. Decreasing on $\left(-\infty,\infty\right)$
• B. Increasing on $\left(-\infty,\infty\right)$
• C. Increasing on $\left(-\infty,0\right)\&\left(\frac85,\infty\right)$, Decreasing on $\left(0,\frac85\right)$
• D. Increasing on $\left(0,\frac85\right)$, Decreasing on $\left(-\infty,0\right)\&\left(\frac85,\infty\right)$

Answer: (D)

Identify the intervals on which the function is increasing or decreasing.
$f(x)=\sin^2x$ on $[0,\pi]$

• A. Increasing on $[0,π]$
• B. Decreasing on $[0,π]$
• C. Increasing on $[0,\frac{\pi}{2})$, Decreasing on $(\frac{\pi}{2},\pi]$
• D. Increasing on $(\frac{\pi}{2},\pi]$, Decreasing on $[0,\frac{\pi}{2})$

Answer: (C)

Determining Where a Function is Increasing & Decreasing Example 1

To determine the intervals where a function f(x) is increasing or decreasing, we analyze the graph of its derivative, f'(x). The key concept is that the sign of the derivative indicates the behavior of the original function: when f'(x) is positive, f(x) is increasing, and when f'(x) is negative, f(x) is decreasing.

Starting from the left, we observe the behavior of f'(x) across different intervals:

1. From negative infinity to -5, the graph of f'(x) is below the x-axis, indicating that f(x) is decreasing in the interval (-∞, -5).

2. At -5, the graph crosses the x-axis and becomes positive, which means f(x) starts increasing from (-5, 3) as f'(x) remains above the x-axis until 3.

3. At 3, f'(x) dips back below the x-axis, indicating that f(x) is decreasing again in the interval (3, 5).

4. Finally, at 5, f'(x) returns to being positive, which means f(x) is increasing once more from (5, ∞).

In summary, the function f(x) is:

• Decreasing on the interval (-∞, -5)
• Increasing on the interval (-5, 3)
• Decreasing on the interval (3, 5)
• Increasing on the interval (5, ∞)

Understanding the relationship between a function and its derivative is crucial for analyzing the behavior of the function based on the sign of the derivative graph.

The First Derivative Test: Finding Local Extrema

Understanding the behavior of a function is crucial in calculus, particularly when identifying local extrema, which are the local maximum and minimum values of a function. The first derivative of a function provides valuable insights into its increasing and decreasing behavior. Specifically, the sign of the derivative indicates whether the function is rising or falling. When the sign of the derivative changes, it signals the presence of local extrema.

The first derivative test is a systematic method used to find these local extrema. To apply this test, one must first identify the critical points of the function, which occur where the derivative is either equal to zero or does not exist. For example, consider the function \( f(x) = x^3 - 3x^2 + 4 \). The first derivative, calculated as \( f'(x) = 3x^2 - 6x \), can be factored to \( f'(x) = 3x(x - 2) \). Setting this equal to zero reveals the critical points at \( x = 0 \) and \( x = 2 \).

Next, a sign chart is created by dividing the number line into intervals based on the critical points. For the function above, the intervals are \( (-\infty, 0) \), \( (0, 2) \), and \( (2, \infty) \). By selecting test values from each interval, one can determine the sign of the derivative in those intervals. For instance, testing \( x = -1 \), \( x = 1 \), and \( x = 3 \) yields positive, negative, and positive results, respectively. This indicates that the function is increasing on \( (-\infty, 0) \), decreasing on \( (0, 2) \), and increasing again on \( (2, \infty) \).

Applying the first derivative test, we observe that the sign changes from positive to negative at \( x = 0 \), indicating a local maximum at this point. Conversely, the sign changes from negative to positive at \( x = 2 \), indicating a local minimum. Thus, the function has a local maximum at \( x = 0 \) and a local minimum at \( x = 2 \).

To find the actual values of these extrema, one would substitute the critical points back into the original function. However, the primary goal of the first derivative test is to identify the locations of these extrema based on the behavior of the derivative. This method is a powerful tool in calculus for analyzing the characteristics of functions and understanding their graphical representations.

The First Derivative Test: Finding Local Extrema Example 2

To identify local minimum and maximum values of the function \( f(x) = (x + 7)^3 \), we begin by applying the first derivative test. This involves finding the critical points where the derivative is either zero or does not exist.

First, we calculate the derivative of the function. Using the power rule, we find:

\[f'(x) = 3(x + 7)^2 \cdot (1) = 3(x + 7)^2\]

Next, we set the derivative equal to zero to find critical points:

\[3(x + 7)^2 = 0\]

Dividing both sides by 3 gives:

\[(x + 7)^2 = 0\]

Taking the square root results in:

\[x + 7 = 0 \implies x = -7\]

Now, we create a sign chart to analyze the intervals around the critical point \( x = -7 \). The intervals are \( (-\infty, -7) \) and \( (-7, \infty) \). We will select test values from each interval to determine the sign of the derivative.

For the interval \( (-\infty, -7) \), we choose \( x = -8 \):

\[f'(-8) = 3(-8 + 7)^2 = 3(-1)^2 = 3 \quad (\text{positive})\]

For the interval \( (-7, \infty) \), we choose \( x = 0 \):

\[f'(0) = 3(0 + 7)^2 = 3(7)^2 = 3 \cdot 49 = 147 \quad (\text{positive})\]

Since the derivative \( f' \) is positive in both intervals, it indicates that the function is increasing on both sides of the critical point \( x = -7 \). Therefore, the sign of the derivative does not change at this critical point, which implies that there are no local extrema.

This conclusion aligns with the nature of cubic functions, which typically exhibit continuous increasing or decreasing behavior without local maxima or minima unless they have specific inflection points. In this case, the function \( f(x) = (x + 7)^3 \) is increasing throughout its domain, confirming that there are no local minimum or maximum values.

The First Derivative Test: Finding Local Extrema Example 3

To find the local minimum and maximum values of the function \( h(t) = \frac{t^3}{2t + 1} \), we start by determining the critical points through the first derivative test. The critical points occur where the first derivative \( h'(t) \) is either equal to zero or does not exist.

Using the quotient rule, we calculate the derivative as follows:

\[h'(t) = \frac{(2t + 1)(3t^2) - (t^3)(2)}{(2t + 1)^2}\]

After simplifying, we find:

\[h'(t) = \frac{4t^3 + 3t^2}{(2t + 1)^2}\]

Next, we set the numerator equal to zero to find critical points:

\[4t^3 + 3t^2 = 0\]

Factoring gives us:

\[t^2(4t + 3) = 0\]

This results in critical points at \( t = 0 \) and \( t = -\frac{3}{4} \). We also check where the derivative does not exist by setting the denominator equal to zero:

\[(2t + 1)^2 = 0 \implies 2t + 1 = 0 \implies t = -\frac{1}{2}\]

However, since the function \( h(t) \) is undefined at \( t = -\frac{1}{2} \), we discard this point as a local extremum.

Now, we create a sign chart using the critical points \( t = -\frac{3}{4} \) and \( t = 0 \). The intervals to test are \( (-\infty, -\frac{3}{4}) \), \( (-\frac{3}{4}, 0) \), and \( (0, \infty) \). We select test points from each interval:

• For \( t = -1 \) (in \( (-\infty, -\frac{3}{4}) \)): \[h'(-1) = \frac{4(-1)^3 + 3(-1)^2}{(2(-1) + 1)^2} = \frac{-4 + 3}{1} = -1 \quad (\text{negative})\]
• For \( t = -\frac{1}{4} \) (in \( (-\frac{3}{4}, 0) \)): \[h'(-\frac{1}{4}) = \frac{4(-\frac{1}{4})^3 + 3(-\frac{1}{4})^2}{(2(-\frac{1}{4}) + 1)^2} = \frac{-\frac{1}{16} + \frac{3}{16}}{\frac{1}{4}} = \frac{\frac{2}{16}}{\frac{1}{4}} = 2 \quad (\text{positive})\]
• For \( t = 1 \) (in \( (0, \infty) \)): \[h'(1) = \frac{4(1)^3 + 3(1)^2}{(2(1) + 1)^2} = \frac{4 + 3}{9} = \frac{7}{9} \quad (\text{positive})\]

From the sign chart, we observe that \( h'(t) \) changes from negative to positive at \( t = -\frac{3}{4} \), indicating a local minimum. At \( t = 0 \), there is no sign change, so it is not a local extremum.

To find the local minimum value, we evaluate \( h \) at the critical point \( t = -\frac{3}{4} \):

\[h\left(-\frac{3}{4}\right) = \frac{\left(-\frac{3}{4}\right)^3}{2\left(-\frac{3}{4}\right) + 1} = \frac{-\frac{27}{64}}{-\frac{6}{4} + \frac{4}{4}} = \frac{-\frac{27}{64}}{-\frac{2}{4}} = \frac{-\frac{27}{64}}{-\frac{1}{2}} = \frac{-27}{32}\]

Thus, the function has a local minimum value of \( -\frac{27}{32} \) at \( t = -\frac{3}{4} \). This can also be expressed as the ordered pair \( \left(-\frac{3}{4}, -\frac{27}{32}\right) \).

Identify the local minimum and maximum values of the given function, if any.
$f\left(\theta\right)=\sin\theta+\cos^2\theta$ on $[0,π]$

• A. Local maxima of $\frac54$ at $x=\frac{\pi}{6}$, Local minimum of $1$ at $x=\frac{\pi}{2}$
• B. Local maxima of $\frac54$ at $x=\frac{\pi}{6}\&\frac{5\pi}{6}$, Local minimum of $1$ at $x=\frac{\pi}{2}$
• C. Local maxima of $\frac54$ at $x=\frac{\pi}{6}$, Local minimum of $1$ at $x=\frac{\pi}{2}\&\frac{3\pi}{2}$
• D. No local extrema

Answer: (B)

The First Derivative Test: Finding Local Extrema Example 4

To identify the local and global extrema of the function \( f(x) = x^2 + 4x + 17 \), we start by recognizing that this function represents a parabola. Since we are not given a closed interval, we will focus on finding the local extrema first and then determine the global extrema based on the characteristics of the function.

We begin by finding the critical points using the first derivative test. The first step is to compute the derivative of the function. Applying the power rule, we find:

\( f'(x) = 2x + 4 \)

Next, we set the derivative equal to zero to find the critical points:

\( 2x + 4 = 0 \)

\( 2x = -4 \)

\( x = -2 \)

Now, we have one critical point at \( x = -2 \). To analyze the behavior of the function around this critical point, we create a sign chart by testing values from the intervals \( (-\infty, -2) \) and \( (-2, \infty) \). We choose test points, such as \( x = -3 \) from the first interval and \( x = 0 \) from the second interval.

Calculating the derivative at these points:

\( f'(-3) = 2(-3) + 4 = -6 + 4 = -2 \) (negative)

\( f'(0) = 2(0) + 4 = 0 + 4 = 4 \) (positive)

From this analysis, we see that the derivative is negative on the interval \( (-\infty, -2) \), indicating that the function is decreasing. Conversely, the derivative is positive on the interval \( (-2, \infty) \), indicating that the function is increasing. Therefore, at the critical point \( x = -2 \), we have a local minimum.

To find the value of this local minimum, we substitute \( x = -2 \) back into the original function:

\( f(-2) = (-2)^2 + 4(-2) + 17 \)

\( = 4 - 8 + 17 \)

\( = 13 \)

Thus, the local minimum value is \( 13 \) at \( x = -2 \).

Next, we consider the global extrema. Since the function \( f(x) = x^2 + 4x + 17 \) is a parabola that opens upwards (as indicated by the positive coefficient of \( x^2 \)), the local minimum we found at \( x = -2 \) is also the global minimum. As the function approaches infinity on both sides, there are no global maximum values.

In conclusion, the function has a local and global minimum value of \( 13 \) at \( x = -2 \). This understanding of the function's behavior reinforces the importance of analyzing both the derivative and the overall shape of the graph when determining extrema.

Here’s what students ask on this topic:

What is the first derivative test and how is it used to find local extrema?

The first derivative test is a method used to determine the local maxima and minima of a function by analyzing the sign changes of its first derivative around critical points. Critical points occur where the derivative is zero or does not exist. To use the test, follow these steps: 1) Find the critical points by setting the derivative equal to zero or identifying where it does not exist. 2) Create intervals around these critical points. 3) Choose test values within each interval and plug them into the derivative. 4) Observe the sign changes: if the derivative changes from positive to negative, there is a local maximum; if it changes from negative to positive, there is a local minimum. This test helps identify the peaks and valleys of the function.

How do you determine where a function is increasing or decreasing using its derivative?

To determine where a function is increasing or decreasing, analyze its first derivative. Follow these steps: 1) Find the derivative of the function. 2) Identify the critical points by setting the derivative equal to zero or finding where it does not exist. 3) Create intervals around these critical points. 4) Choose test values within each interval and plug them into the derivative. 5) Observe the sign of the derivative: if the derivative is positive, the function is increasing; if the derivative is negative, the function is decreasing. This method helps understand the behavior of the function over different intervals.

What are critical points and how do you find them?

Critical points are points on a function where the derivative is either zero or does not exist. These points are important for determining where a function changes direction, indicating potential local maxima or minima. To find critical points, follow these steps: 1) Take the derivative of the function. 2) Set the derivative equal to zero and solve for the variable. 3) Identify any points where the derivative does not exist. The solutions to these steps are the critical points of the function.

How do you use a sign chart to determine intervals of increase and decrease?

A sign chart helps determine intervals where a function is increasing or decreasing by analyzing the sign of its derivative. Follow these steps: 1) Find the critical points by setting the derivative equal to zero or identifying where it does not exist. 2) Create a number line and mark the critical points. 3) Divide the number line into intervals based on these points. 4) Choose test values within each interval and plug them into the derivative. 5) Observe the sign of the derivative: if positive, the function is increasing; if negative, the function is decreasing. This method provides a clear visual representation of the function's behavior over different intervals.

What is the significance of the sign of the first derivative in determining the behavior of a function?

The sign of the first derivative is crucial in determining the behavior of a function. If the first derivative is positive, the function is increasing, meaning the function's values are rising as the input increases. Conversely, if the first derivative is negative, the function is decreasing, meaning the function's values are falling as the input increases. By analyzing the sign of the first derivative, we can identify intervals of increase and decrease, as well as locate local maxima and minima using the first derivative test. This analysis helps understand the overall shape and behavior of the function.