Problem 123 - Thomas Calculus

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? \begin{equation}y=\frac{4}{5} x^{5}+16 x^{2}-25\end{equation}

Step-by-Step Solution

Answer

Local max at (-2, 13.4), local min at (0, -25), inflection near (-1.26, y-value).

Step 1: Find the first derivative

To determine critical points and behavior of the function, we first find the derivative of the function using power rules. Given \[y = \frac{4}{5}x^5 + 16x^2 - 25\]The first derivative, \(y'\), is:\[y' = \frac{d}{dx}\left(\frac{4}{5}x^5 + 16x^2 - 25\right) = 4x^4 + 32x\]

Step 2: Find critical points

Critical points occur where the first derivative is zero or undefined. Here, we set the first derivative equal to zero:\[4x^4 + 32x = 0\]Factor the expression:\[4x(x^3 + 8) = 0\]This gives solutions:\(x = 0\) \ \(x = -2\) (since \(x^3 + 8 = (x + 2)(x^2 - 2x + 4) = 0\) only provides a real solution at \(x = -2\)).

Step 3: Find the second derivative

The second derivative helps identify inflection points. Differentiate the first derivative:\[y'' = \frac{d}{dx}(4x^4 + 32x) = 16x^3 + 32\]

Step 4: Evaluate inflection points

Inflection points occur where the second derivative changes sign, which happens where it is zero. Solve:\[16x^3 + 32 = 0\] Solve for \(x\):\[16x^3 = -32\] \[x^3 = -2\] \(x = -\sqrt[3]{2}\) \(x \approx -1.26\) is the x-coordinate for the inflection point.

Step 5: Determine behavior at critical points

Take the second derivative to test the sign around critical points \(x = 0\) and \(x = -2\). Utilize the second derivative test: - For \(x = 0\), \(y'' = 16(0)^3 + 32 = 32 > 0\): Local minimum.- For \(x = -2\), \(y'' = 16(-2)^3 + 32 = -128 + 32 = -96 < 0\): Local maximum.

Step 6: Calculate the coordinates of the critical and inflection points

Plug the critical and inflection points' x-values back into the original function to find their y-coordinates:- At \(x = 0\): \(y = \frac{4}{5}(0)^5 + 16(0)^2 - 25 = -25\)- At \(x = -2\): \(y = \frac{4}{5}(-2)^5 + 16(-2)^2 - 25\)\(= -\frac{4}{5} \times 32 + 64 - 25\)\( = -25.6 + 64 - 25 = 13.4\)- At \(x \approx -1.26\) for inflection:\(y \approx \frac{4}{5}(-1.26)^5 + 16(-1.26)^2 - 25\) solve for approximate value.

Step 7: Interpret graphs of function and its derivatives

The first derivative graph intersects the x-axis at critical points \(x = 0\) and \(x = -2\), indicating slope changes from positive to negative or vice versa at these points. The second derivative graph intersects the x-axis at \(x \approx -1.26\), indicating the point of inflection. Each intersection marks a change in behavior in the graph of the function.

Key Concepts

Critical Points

In calculus, critical points are where a function's derivative is zero or undefined. These points are significant because they can indicate potential locations of local maxima or minima.
Critical points occur when the first derivative of the function, denoted as \(y'\), equals zero. These points are also essential when analyzing the curvature and direction changes of the graph.

• To find the critical points of the function \(y = \frac{4}{5}x^5 + 16x^2 - 25\), we first found its first derivative: \(y' = 4x^4 + 32x\).
• Setting the first derivative to zero, \(4x^4 + 32x = 0\), we factored it as \(4x(x^3 + 8) = 0\) yielding \(x = 0\) and \(x = -2\) as critical points.

These points are crucial as they are further investigated to classify them using the second derivative.

First Derivative

The first derivative of a function provides the slope of the tangent line to the curve at any point. It offers valuable information about the rate of change of the function.
The derivative, often denoted as \(y'\), also reveals whether a function is increasing or decreasing:

• If \(y' > 0\), the function is increasing.
• If \(y' < 0\), the function is decreasing.

The relation of the first derivative to the original function in this exercise is:
Given \(y = \frac{4}{5}x^5 + 16x^2 - 25\), the first derivative is \(y' = 4x^4 + 32x\).
By analyzing this derivative at various points, we can determine the behavior of the original function concerning its rising or falling slopes, aiding in the identification of critical points.

Second Derivative

The second derivative of a function tells us about the concavity of the graph and provides insight into the points of inflection. It is a powerful tool in understanding the behavior and shape of the graph.
The second derivative is obtained by differentiating the first derivative. For the function analyzed, the second derivative is \(y'' = 16x^3 + 32\).

• If \(y'' > 0\), the graph is concave up, resembling a "U" shape.
• If \(y'' < 0\), the graph is concave down, resembling an "n" shape.

This information helps identify inflection points, where the graph changes concavity. In this example, the point where \(y'' = 0\) marks a potential point of inflection, occurring at \(x \approx -1.26\). This point marks where the curve transitions from being concave down to concave up.

Local Maximum

A local maximum of a function is a point where the function reaches a peak within a region. It is found where the first derivative changes from positive to negative.
Using the second derivative test, we can confirm the nature of these points:

• For the critical point \(x = -2\), the second derivative \(y'' = 16(-2)^3 + 32 = -96\), which is less than zero. This indicates that \(x = -2\) is a local maximum because the graph changes from concave up to concave down, and therefore reaches a peak here.

To find the coordinates of the local maximum, we place \(x = -2\) back into the original function to get the corresponding \(y\)-value. For this function, the local maximum is at the coordinate \((-2, 13.4)\). This is the highest point in the immediate area of the function's graph.

Local Minimum

A local minimum is a point on a function where the function value is lower than all other surrounding values. It is identified through the first derivative changing from negative to positive.
The second derivative test helps in classifying critical points as local minima:

• At \(x = 0\), \(y'' = 16(0)^3 + 32 = 32\), which is greater than zero.

This indicates that \(x = 0\) is a local minimum, as the function changes from concave down to concave up at this point, indicating a trough in the graph.
Plugging \(x = 0\) back into the original function yields the point \((0, -25)\) as the coordinate of the local minimum. This value represents a low point, establishing the base in the surrounding region of the function's graph.