Problem 115 - Thomas Calculus in SI Units

Question

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 simultaneously. 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? $$y=\frac{4}{5} x^{5}+16 x^{2}-25$$

Step-by-Step Solution

Answer

Local maximum at \((-2, \frac{47}{5})\), local minimum at \((0, -25)\), inflection at \(x = -\sqrt[3]{2}\).

Step 1: Differentiate the Function

First, we find the first derivative of the function \( y = \frac{4}{5} x^{5} + 16x^{2} - 25 \). The derivative is \( y' = \frac{d}{dx}(\frac{4}{5} x^{5} + 16x^{2} - 25) = 4x^4 + 32x \).

Step 2: Find Critical Points

Set the first derivative equal to zero to find the critical points: \( 4x^4 + 32x = 0 \). Factor the equation: \( 4x(x^3 + 8) = 0 \). The solutions are \( x = 0 \) and \( x^3 = -8 \), so \( x = -2 \). Thus, the critical points are at \( x = 0 \) and \( x = -2 \).

Step 3: Differentiate Again for Concavity

Now, find the second derivative to determine concavity: \( y'' = \frac{d}{dx}(4x^4 + 32x) = 16x^3 + 32 \).

Step 4: Determine Inflection Points

Set the second derivative equal to zero to find potential inflection points: \( 16x^3 + 32 = 0 \). Simplify: \( x^3 = -2 \), so \( x = -\sqrt[3]{2} \). Verify sign changes around this point to confirm it as an inflection point.

Step 5: Evaluate Function at Critical Points

Evaluate the original function \( y = \frac{4}{5} x^{5} + 16x^{2} - 25 \) at the critical points:- At \( x = 0 \), \( y(0) = -25 \).- At \( x = -2 \), \( y(-2) = \frac{4}{5}(-2)^5 + 16(-2)^2 - 25 = -\frac{128}{5} + 64 - 25 = \frac{47}{5} \).

Step 6: Analyze the Derivatives

Determine how the zeros of the derivatives relate to the original function:- The zeros of \( y' \) (at \( x = 0 \) and \( x = -2 \)) correspond to potential maxima and minima of the original function.- The zero of \( y'' \) (at \( x = -\sqrt[3]{2} \)) indicates a possible inflection point on the graph of the function.

Step 7: Graph the Original and Derivative Functions

Graph the original function, first derivative \( y' \), and second derivative \( y'' \). Ensure the graph shows critical points and the inflection point:- Highlight \( (0, -25) \) as a local minimum.- Highlight \( (-2, \frac{47}{5}) \) as a local maximum.- Mark \( x = -\sqrt[3]{2} \) as an inflection point.

Step 8: Relationship of Derivatives to Function

The zeros of the first derivative \( y' \) (tangent line horizontal points) show potential maxima and minima of the function. The change in sign of the second derivative \( y'' \) is indicative of inflection points where concavity changes. Derivatives provide the slope behavior and concavity details needed for sketching the overall shape of the original function.

Key Concepts

Inflection Points

Inflection points are where the graph of a function changes concavity from concave up (like a smile) to concave down (like a frown) or vice versa. To find inflection points, we use the second derivative of the function. In this exercise, the second derivative is given by \( y'' = 16x^3 + 32 \).

• Set this equation to zero: \( 16x^3 + 32 = 0 \).
• Solve for \( x \) to get potential inflection points. For this function, \( x = -\sqrt[3]{2} \).
• Check intervals around this \( x \) value to see if the concavity changes. If it does, \( x = -\sqrt[3]{2} \) is indeed an inflection point.

Calculating the value of the function at this \( x \) can further help in graphing tasks. Keep in mind that an inflection point does not necessarily imply a maximum or minimum, just a change in the shape of the curve.

Critical Points

Critical points occur where the first derivative of a function is zero or undefined. They are important because they can indicate where a function might have relative maxima or minima. For the function \( y = \frac{4}{5} x^{5} + 16x^{2} - 25 \), the first derivative is \( y' = 4x^4 + 32x \).

• Set this derivative equal to zero: \( 4x(x^3 + 8) = 0 \).
• Solve to find \( x = 0 \) and \( x = -2 \).
• These \( x \) values are the critical points. Check the behavior of the second derivative at these points to determine if they are maxima or minima.

Evaluating the original function at these critical points, we find the function's values, helping us understand the local high and low points on the graph. In this case, \( (0, -25) \) is a local minimum and \( (-2, \frac{47}{5}) \) is a local maximum.

Second Derivative Test

The Second Derivative Test is a convenient method to determine whether a critical point is a local maximum, local minimum, or neither. Once a function's critical points are known, the second derivative helps to assess concavity:

• Calculate the second derivative: For our function, it is \( y'' = 16x^3 + 32 \).
• Evaluate the second derivative at each critical point.
• If \( y''(x) > 0 \), the function is concave up at \( x \), indicating a local minimum.
• If \( y''(x) < 0 \), the function is concave down at \( x \), suggesting a local maximum.
• If \( y''(x) = 0 \), the test is inconclusive.

In the exercise, testing at \( x = 0 \) and \( x = -2 \) identified which were maxima and minima. This insight allows us to sketch the function more accurately.

Graphing Functions

Graphing a function and its derivatives provides a visual understanding of mathematical behavior. In this exercise, graphing helps us verify our findings:

• Plot the original function \( y = \frac{4}{5} x^{5} + 16x^{2} - 25 \) over a suitable domain.
• Include critical points \( (0, -25) \) and \( (-2, \frac{47}{5}) \), and the inflection point \( x = -\sqrt[3]{2} \).
• Overlay graphs of the first and second derivatives, \( y' \) and \( y'' \), to visually indicate where the function is increasing, decreasing, concave up, or concave down.
• Observe intersections with the \( x \)-axis. The zeros in \( y' \) (critical points) suggest changes in slope directions of the original function.
• The zeros in \( y'' \) give places of concavity change.

Graphing confirms analytical work and highlights the relationships between a function and its derivatives, emphasizing where the function's behavior alters.