Problem 123 - University Calculus: Early Transcendentals

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? $$y=x^{5}-5 x^{4}-240$$

Step-by-Step Solution

Answer

Local minimum at \((4, -496)\); inflection points at \((0, -240)\) and \((3, -402)\).

Step 1: Find the First Derivative

To find the critical points where the function yields local maxima or minima, we first need the first derivative of the function. For the given function \( y = x^5 - 5x^4 - 240 \), the first derivative \( y' \) is calculated as follows:\[ y' = \frac{d}{dx}(x^5 - 5x^4 - 240) = 5x^4 - 20x^3 \]

Step 2: Find Critical Points

Set the first derivative equal to zero to find critical points:\[ 5x^4 - 20x^3 = 0 \]Factor out the greatest common factor:\[ 5x^3(x - 4) = 0 \]Solve for \( x \):\[ x^3 = 0 \quad \text{or} \quad x - 4 = 0 \]\[ x = 0 \quad \text{or} \quad x = 4 \]

Step 3: Determine the Nature of Critical Points

To classify the critical points found in step 2, use the second derivative test. First, find the second derivative of the function:\[ y'' = \frac{d}{dx}(5x^4 - 20x^3) = 20x^3 - 60x^2 \]Evaluate \( y'' \) at the critical points:- For \( x = 0 \): \( y''(0) = 20(0)^3 - 60(0)^2 = 0 \)- For \( x = 4 \): \( y''(4) = 20(4)^3 - 60(4)^2 = 1280 - 960 = 320 \) Since \( y''(0) = 0 \), use a different method to test if it's a maximum, minimum, or neither. However, \( y''(4) > 0 \) implies a local minimum at \( x = 4 \).

Step 4: Check Second Derivative for Inflection Points

To find inflection points, we need to solve \( y'' = 0 \):\[ 20x^3 - 60x^2 = 0 \]Factor out the common terms:\[ 20x^2(x - 3) = 0 \]Solve for \( x \):\[ x^2 = 0 \quad \text{or} \quad x - 3 = 0 \]\[ x = 0 \quad \text{or} \quad x = 3 \]An inflection point is only valid if there is a sign change in the second derivative at these points.

Step 5: Evaluate the Function at Key Points

Calculate the function value at the potential points from the second derivative test and critical points to ascertain local maximum, minimum, or inflection:- At \( x = 0 \), \( y = 0^5 - 5(0)^4 - 240 = -240 \)- At \( x = 3 \), \( y = 3^5 - 5(3)^4 - 240 = 243 - 405 - 240 = -402 \)- At \( x = 4 \), \( y = 4^5 - 5(4)^4 - 240 = 1024 - 1280 - 240 = -496 \)Thus, the coordinates are:- Local minimum at \((4, -496)\)- Inflection points around \((0, -240)\) and \((3, -402)\)

Step 6: Summary and Graphical Analysis

The function has a local minimum at \((4, -496)\), and possible inflection points at \((0, -240)\) and \((3, -402)\). Graph the function and its first and second derivatives. The points where the first derivative intersects the x-axis correspond to the critical points, while the intersections of the second derivative correspond to the candidate inflection points.

Key Concepts

Inflection Points

Inflection points play a crucial role in understanding how a graph behaves. They are points on the curve where the concavity changes. This means that the graph switches from being "concave up" (shaped like a cup) to "concave down" (shaped like a cap), or vice versa.

To find inflection points in a function, we need to analyze the second derivative. Specifically, we set the second derivative equal to zero and solve, which identifies possible inflection points. However, a valid inflection point must exhibit a change in concavity – there must be a sign change in the second derivative as we pass through that point.

In our exercise, when evaluating the function's second derivative, we found \[ y'' = 20x^3 - 60x^2 = 0 \]Solving, we get potential inflection points at \( x = 0 \) and \( x = 3 \). To confirm these are real inflection points, verify that \( y''(x) \) changes sign around these values.

Critical Points

Critical points of a function are extremely important in calculus as they help us identify where the function reaches its highest or lowest values – its local maxima and minima. A critical point occurs at any \( x \) where the first derivative \( y' \) is zero or undefined.

For the function in question, \[ y' = 5x^4 - 20x^3 \]we set \( y' = 0 \) to find critical points. This yields the points \( x = 0 \) and \( x = 4 \).

To determine whether these critical points represent maxima, minima, or neither, we have to use the second derivative test. By doing this on our critical points, we found that \( x = 4 \) corresponds to a local minimum because \( y''(4) > 0 \). However, since \( y''(0) = 0 \), further investigation is needed at \( x = 0 \) to determine the nature of that critical point.

First Derivative

The first derivative of a function, often denoted \( y' \), is key in identifying critical points along a graph. It tells us where the function increases or decreases, by representing the slope of the tangent line to the curve at any given point.

When the first derivative is zero, it indicates a potential point where the function's slope is horizontal, suggesting either a peak (maximum) or a trough (minimum), known as critical points. For \( y = x^5 - 5x^4 - 240 \), the first derivative is obtained by differentiating:\[ y' = 5x^4 - 20x^3 \].
By setting this equal to zero, \( 5x^4 - 20x^3 = 0 \),we find our critical points: \( x = 0 \) and \( x = 4 \). These points are then further analyzed with the second derivative to accurately classify their nature.

Second Derivative

The second derivative of a function gives us insight into the concavity of the function's graph. It helps confirm whether a critical point is a local maximum, local minimum, or neither, and also aids in identifying inflection points through changes in concavity.

For our function, the second derivative is calculated as:\[ y'' = 20x^3 - 60x^2 \].
This derivative aids in determining concavity: if \( y'' > 0 \), the graph is concave up; if \( y'' < 0 \), the graph is concave down.

Additionally, by setting the second derivative equal to zero:\[ 20x^3 - 60x^2 = 0 \], we find potential inflection points at \( x = 0 \) and \( x = 3 \). The validation step involves checking for changes in sign around these values.Understanding the second derivative and its importance in confirming and classifying critical and inflection points is essential for mastering calculus.