Problem 85 - Thomas Calculus

Question

In Exercises 85–88, find the inflection points (if any) on the graph of the function and the coordinates of the points on the graph where the function has a local maximum or local minimum value. Then graph the function in a region large enough to show all these points simulta0neously. Add to your picture the graphs of the function’s first and second derivatives. How are the values at which these graphs intersect the \(x\) -axis related to the graph of the function? In what other ways are the graphs of the derivatives related to the graph of the function? $$ y=x^{5}-5 x^{4}-240 $$

Step-by-Step Solution

Answer

Inflection Points: \( x = 0, 3 \). Local Minima: \( x = 4 \). Critical Points: \( x = 0, 4 \).

Step 1: Find the First Derivative

First, we need to find the first derivative of the function, which is denoted as \( y' \), to identify critical points. The given function is \( y = x^5 - 5x^4 - 240 \). Using basic differentiation rules, the first derivative is calculated as follows:\[ y' = \frac{d}{dx}(x^5) - \frac{d}{dx}(5x^4) - 0 = 5x^4 - 20x^3. \]

Step 2: Find Critical Points

To find the critical points, solve for \( x \) where \( y' = 0 \). This requires setting the derivative equal to zero and solving the equation:\[ 5x^4 - 20x^3 = 0. \]Factoring out \( 5x^3 \) gives:\[ 5x^3(x - 4) = 0. \]Set each factor equal to zero:1. \( 5x^3 = 0 \Rightarrow x = 0 \).2. \( x - 4 = 0 \Rightarrow x = 4 \).Thus, the critical points are \( x = 0 \) and \( x = 4 \).

Step 3: Determine Local Maxima and Minima

Use the second derivative test to determine if these critical points are maxima or minima. First, find the second derivative:\[ y'' = \frac{d}{dx}(5x^4 - 20x^3) = 20x^3 - 60x^2. \]Evaluate the second derivative at the critical points:1. \( y''(0) = 20(0)^3 - 60(0)^2 = 0 \) (the test is inconclusive here).2. \( y''(4) = 20(4)^3 - 60(4)^2 = 1280 - 960 = 320 \).Since \( y''(4) > 0 \), \( x = 4 \) is a point of local minimum.

Step 4: Identify Inflection Points

To identify inflection points, solve for \( x \) where the second derivative is zero:\[ 20x^3 - 60x^2 = 0. \]Factor out \( 20x^2 \):\[ 20x^2(x - 3) = 0. \]Set each factor to zero:1. \( 20x^2 = 0 \Rightarrow x = 0 \).2. \( x - 3 = 0 \Rightarrow x = 3 \).These points \( x = 0 \) and \( x = 3 \) are potential inflection points. Verify by checking the change of sign in \( y'' \) around these points.

Step 5: Graph the Function and Derivatives

Graph the original function \( y = x^5 - 5x^4 - 240 \) and its derivatives \( y' = 5x^4 - 20x^3 \) and \( y'' = 20x^3 - 60x^2 \). Points where \( y' \) crosses the \( x \)-axis are the critical points, and points where \( y'' \) crosses the \( x \)-axis are potential inflection points. Observe how these affect the graph's curvature and slope.

Key Concepts

Critical Points

Critical points are essential in understanding where a function exhibits a change in its slope or gradient. These points occur where the derivative of the function, denoted as \( y' \), is equal to zero. In simpler terms, the tangent to the curve is horizontal at these points.

To find critical points, calculate the first derivative of the function and solve the equation \( y' = 0 \). This procedure identifies potential locations for local maxima, minima, or saddle points. In our example, the derivative is:

• \( y' = 5x^4 - 20x^3 \)

By factoring, we arrived at \( 5x^3(x - 4) = 0 \), which gives us critical points at \( x = 0 \) and \( x = 4 \).

Understanding these points plays a significant role in sketching the graph's behavior and further analyzing the function's characteristics.

Derivatives

Derivatives are vital tools in calculus that provide insights into the rate of change of a function. The first derivative, \( y' \), indicates how the function's output changes with respect to small changes in input. It essentially tells us how the slope of the tangent to a curve behaves.

For the function \( y = x^5 - 5x^4 - 240 \), we computed:

• \( y' = 5x^4 - 20x^3 \)
• \( y'' = 20x^3 - 60x^2 \)

The second derivative, \( y'' \), offers additional insight into the curvature of the function. It helps us determine whether critical points correspond to local maxima, minima, or points of inflection by considering the concavity of the function.

In practice:

• Positive \( y'' \) indicates a local minimum (concave up)
• Negative \( y'' \) indicates a local maximum (concave down)
• Zero \( y'' \) leaves us needing further investigation

Local Maxima and Minima

Local maxima and minima are points where the function reaches a highest or lowest value, at least within its immediate vicinity. These concepts help us understand where the function is peaking or dipping.

After identifying critical points, we apply the second derivative test:

• If \( y'' > 0 \) at a critical point, it's a local minimum, indicating a small u-shaped section of the graph.
• If \( y'' < 0 \), then it's a local maximum, resembling an upside-down u shape.
• If \( y'' = 0 \), the point needs further analysis, as it might neither be a maximum nor a minimum.

Using the second derivative, \( y'' = 20x^3 - 60x^2 \), we evaluated it at our critical points:

• At \( x = 4 \), \( y''(4) = 320 \) (positive), signifying a local minimum
• At \( x = 0 \), \( y''(0) = 0 \), which remains inconclusive within this test

Graphing Functions

Graphing functions allows us to visualize and better understand the behavior of mathematical models. For our function, graphing helps us identify critical points, local extrema, and inflection points.

When graphing:

• Plot the original function \( y \), as well as its first \( y' \) and second derivatives \( y'' \).
• The graph of \( y' \) shows where the function's gradient is zero (critical points), as it intersects the x-axis here.
• The graph of \( y'' \) intersects the x-axis at potential inflection points, where the concavity changes.

Visual representation solidifies the abstract concepts of derivatives and critical points, showcasing:

• Curves' symmetry and overall shape.
• Peaks and troughs identified by the second derivative test.
• Transitions in curvature that indicate inflection points.

Graphing is a powerful tool for finding detailed patterns and relationships between the function and its derivatives, making it easier to deduce mathematical properties.