The First Derivative Test: Videos & Practice Problems

Understanding local extrema is crucial in calculus, particularly when analyzing functions without their graphs. Local extrema refer to the highest or lowest points in a specific interval of a function. To identify these points, we first need to determine the function's critical points, which are essential for locating local maxima and minima.

Critical points occur where the derivative of a function is either zero or undefined. This is significant because local extrema typically correspond to horizontal tangent lines (where the derivative equals zero) or points where the derivative does not exist. For example, if we consider a function f(x) = x³ - 12x + 5, we can find its critical points by first calculating its derivative:

$$f'(x) = 3x^2 - 12$$

Next, we set the derivative equal to zero to find where the slope of the tangent line is horizontal:

$$3x^2 - 12 = 0$$

Solving this equation involves adding 12 to both sides:

$$3x^2 = 12$$

Then, dividing both sides by 3 gives:

$$x^2 = 4$$

Taking the square root of both sides results in:

$$x = \pm 2$$

Thus, we have two critical points at x = 2 and x = -2. However, it is important to note that not all critical points correspond to local extrema. For instance, a critical point may have a horizontal tangent but not be a local maximum or minimum. This distinction is vital in further analysis.

In addition to identifying local extrema, critical points can also be used to explore other characteristics of a function, such as concavity and inflection points. As we progress in our studies, we will delve deeper into the applications of critical points and how they can enhance our understanding of various functions.

To find the critical points of the function \( f(x) = \frac{x^2}{x - 4} \), we first need to determine where the derivative \( f'(x) \) is equal to 0 or does not exist. Since this is a rational function, we will apply the quotient rule for differentiation.

The quotient rule states that if you have a function in the form \( \frac{g(x)}{h(x)} \), the derivative is given by:

\[f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{(h(x))^2}\]

In our case, \( g(x) = x^2 \) and \( h(x) = x - 4 \). The derivatives are \( g'(x) = 2x \) and \( h'(x) = 1 \). Applying the quotient rule, we have:

\[f'(x) = \frac{(x - 4)(2x) - (x^2)(1)}{(x - 4)^2}\]

Expanding this gives:

\[f'(x) = \frac{2x^2 - 8x - x^2}{(x - 4)^2} = \frac{x^2 - 8x}{(x - 4)^2}\]

Next, we can factor the numerator:

\[f'(x) = \frac{x(x - 8)}{(x - 4)^2}\]

To find the critical points, we set the numerator equal to 0:

\[x(x - 8) = 0\]

This gives us two solutions:

\[x = 0 \quad \text{and} \quad x = 8\]

Now, we also need to consider where the derivative does not exist. The derivative will be undefined where the denominator is zero, which occurs when:

\[(x - 4)^2 = 0 \quad \Rightarrow \quad x = 4\]

However, we must check if this point is within the domain of the original function. The function \( f(x) \) is undefined at \( x = 4 \) because it would make the denominator zero. Therefore, \( x = 4 \) is not a critical point.

In conclusion, the critical points of the function \( f(x) = \frac{x^2}{x - 4} \) are:

\[x = 0 \quad \text{and} \quad x = 8\]

These points are where the derivative is either zero or does not exist, while also being within the domain of the original function.

To find the critical points of the function \( y = \ln(x) - 1 \), we first need to determine its derivative. The derivative of the natural logarithm function is given by the formula:

\( y' = \frac{1}{x} \)

In this case, since we are looking at \( y = \ln(x) - 1 \), the derivative simplifies to:

\( y' = \frac{1}{x} \)

Next, critical points occur where the derivative is either equal to zero or does not exist. However, for the function \( y' = \frac{1}{x} \), there are no values of \( x \) that make the numerator zero, meaning \( y' \) cannot equal zero. Therefore, we focus on where the derivative does not exist, which happens when the denominator is zero. Setting the denominator equal to zero gives:

\( x = 0 \)

However, we must also consider the domain of the original function \( y = \ln(x) - 1 \). The natural logarithm function is only defined for \( x > 0 \). Thus, \( x = 0 \) is not within the domain of the original function.

Additionally, if we check \( x = 1 \), we find that plugging this value into the original function results in \( \ln(1) - 1 = 0 - 1 = -1 \), which is valid. However, since the derivative \( y' \) does not exist at \( x = 1 \) (as it leads to a division by zero), we must confirm that this point is not included in the domain of the function.

Consequently, since there are no values of \( x \) that yield critical points within the domain of the function \( y = \ln(x) - 1 \), we conclude that there are no critical points for this function.

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 \( x = 0 \):

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

Since \( f'(0) \) is positive, the function is increasing at \( x = 0 \).

For \( x = 5 \):

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

Here, \( f'(5) \) is negative, indicating that the function is decreasing at \( x = 5 \).

While determining behavior at specific points is useful, it is often more common to find entire intervals where the function is increasing or decreasing. This process begins by identifying critical points, which occur where the derivative is zero or undefined. In our case, we set the derivative equal to zero:

\[ -2x + 4 = 0 \]

Solving for \( x \) gives:

\[ x = 2 \]

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

Choosing \( x = 0 \) from the interval \( (-\infty, 2) \):

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

Choosing \( x = 5 \) from the interval \( (2, \infty) \):

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

From this analysis, we conclude that the function is increasing on the interval \( (-\infty, 2) \) and decreasing on the interval \( (2, \infty) \). This method of using critical points and a sign chart allows for a comprehensive understanding of the function's behavior across its entire domain.

In summary, to determine where a function is increasing or decreasing, first find the critical points by setting the derivative to zero. Then, create a sign chart to analyze the intervals formed by these critical points. By testing values within these intervals, you can ascertain whether the function is increasing or decreasing based on the sign of the derivative.

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 \( f'(x) > 0 \), the function \( f(x) \) is increasing, and when \( f'(x) < 0 \), the function \( f(x) \) is decreasing.

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

1. From \( -\infty \) to \( -5 \), the graph of \( f'(x) \) is below the x-axis, indicating that \( f(x) \) is decreasing in this interval.

2. At \( x = -5 \), \( f'(x) \) crosses above the x-axis, becoming positive. This means that \( f(x) \) starts increasing from \( -5 \) to \( 3 \).

3. At \( x = 3 \), \( f'(x) \) dips back below the x-axis, indicating that \( f(x) \) is decreasing again from \( 3 \) to \( 5 \).

4. Finally, at \( x = 5 \), \( f'(x) \) is positive once more, which means \( f(x) \) is increasing from \( 5 \) to \( +\infty \).

In summary, the function \( f(x) \) is:

• Decreasing on the interval \( (-\infty, -5) \)
• Increasing on the interval \( (-5, 3) \)
• Decreasing on the interval \( (3, 5) \)
• Increasing on the interval \( (5, +\infty) \)

When analyzing a graph of a derivative, always focus on the sign of the derivative to determine the behavior of the original function. This approach is essential for understanding the relationship between a function and its derivative.

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 for finding 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 \( (2, \infty) \). A test value from each interval is then substituted into the first derivative to determine the sign of the derivative in that interval. For instance, testing \( x = -1 \) yields a positive derivative, indicating the function is increasing. Testing \( x = 1 \) results in a negative derivative, indicating the function is decreasing, and testing \( x = 3 \) again yields a positive derivative, indicating the function is increasing.

By analyzing the sign changes, one can apply the first derivative test: if the derivative changes from positive to negative at a critical point, a local maximum exists there; conversely, if it changes from negative to positive, a local minimum exists. In our example, 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 local maxima and minima based on the behavior of the derivative. This method is a powerful tool in calculus for understanding the characteristics of functions and their graphs.

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:

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

Dividing both sides by 3 gives:

Taking the square root results in:

Thus, the only critical point is at \( x = -7 \).

To determine the behavior of the function around this critical point, we create a sign chart by dividing the number line into intervals based on the critical point. The intervals are \( (-\infty, -7) \) and \( (-7, \infty) \).

Next, we select test values from each interval to evaluate 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 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 characteristics 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.

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. Critical points occur where the first derivative \( h'(t) \) is either zero or undefined. Since \( h(t) \) is a rational function, we will apply the quotient rule to find \( h'(t) \).

The quotient rule states that if you have a function \( \frac{f(t)}{g(t)} \), then the derivative is given by:

\[h'(t) = \frac{g(t)f'(t) - f(t)g'(t)}{(g(t))^2}\]

For our function, let \( f(t) = t^3 \) and \( g(t) = 2t + 1 \). The derivatives are \( f'(t) = 3t^2 \) and \( g'(t) = 2 \). Applying the quotient rule, we have:

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

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 out \( t^2 \), we get:

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

This gives us critical points at \( t = 0 \) and \( t = -\frac{3}{4} \). We also need to check where the derivative does not exist, which occurs when the denominator is zero:

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

However, since \( t = -\frac{1}{2} \) makes the original function undefined, it cannot be a local extremum. Thus, we focus on the critical points \( t = 0 \) and \( t = -\frac{3}{4} \).

Next, we create a sign chart to analyze the intervals defined by these critical points: \( (-\infty, -\frac{3}{4}) \), \( (-\frac{3}{4}, 0) \), and \( (0, \infty) \). We select test points from each interval to determine the sign of \( h'(t) \).

1. For \( t = -1 \) (in \( (-\infty, -\frac{3}{4}) \)): \[ h'(-1) = \frac{(-1)^2(4(-1) + 3)}{(2(-1) + 1)^2} = \frac{1 \cdot (-1)}{1} < 0 \quad \text{(decreasing)} \] 2. For \( t = -\frac{1}{4} \) (in \( (-\frac{3}{4}, 0) \)): \[ h'(-\frac{1}{4}) = \frac{(-\frac{1}{4})^2(4(-\frac{1}{4}) + 3)}{(2(-\frac{1}{4}) + 1)^2} = \frac{\frac{1}{16} \cdot 1}{\frac{1}{4}} > 0 \quad \text{(increasing)} \]3. For \( t = 1 \) (in \( (0, \infty) \)): \[ h'(1) = \frac{(1)^2(4(1) + 3)}{(2(1) + 1)^2} > 0 \quad \text{(increasing)} \]

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

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 \( \left(-\frac{3}{4}, -\frac{27}{32}\right) \).

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.

To find the local extrema, we apply the first derivative test. We begin by calculating the first derivative of the function using the power rule:

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

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

\( 2x + 4 = 0 \)

\( 2x = -4 \)

\( x = -2 \)

We have identified one critical point at \( x = -2 \). To analyze the behavior of the function around this critical point, we create a sign chart by dividing the number line into intervals based on the critical point. The intervals are \( (-\infty, -2) \) and \( (-2, \infty) \).

We select test values from each interval to determine the sign of the derivative:

• For the interval \( (-\infty, -2) \), we choose \( x = -3 \):

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

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

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

From this analysis, we see that the derivative is negative on the interval \( (-\infty, -2) \) and positive on the interval \( (-2, \infty) \). This indicates that the function is decreasing before \( x = -2 \) and increasing after \( x = -2 \). Therefore, we conclude that there is a local minimum at \( x = -2 \).

To find the value of the 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 \( x = -2 \).

Now, to determine the global extrema, we consider the nature of the parabola. Since the coefficient of \( x^2 \) is positive, the parabola opens upwards. This means that the local minimum we found is also the lowest point on the graph, making it the global minimum as well.

In conclusion, the function \( f(x) = x^2 + 4x + 17 \) has a local and global minimum value of \( 13 \) at \( x = -2 \). The function does not have any local or global maximum values, as it continues to increase indefinitely on both sides of the vertex.