Problem 87 - 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? $$ y=\frac{4}{5} x^{5}+16 x^{2}-25 $$

Step-by-Step Solution

Answer

Critical points at \((0, -25)\) and \((-2,7)\); no clear inflection points found.

Step 1: Find the First Derivative

The first step in finding inflection points and local extrema is to determine the first derivative of the function, as it will give us critical points for the local extrema and is also necessary to find the inflection points. Given function: \[ y = \frac{4}{5} x^5 + 16x^2 - 25 \]Differentiate with respect to \(x\):\[ y' = \frac{d}{dx}\left(\frac{4}{5} x^5 + 16x^2 - 25\right) \]The first derivative is:\[ y' = \left(4x^4 + 32x\right) \]

Step 2: Find Critical Points

To find the critical points, we need to set the first derivative equal to zero and solve for \(x\):\[ 4x^4 + 32x = 0 \]Factor out a common term:\[ 4x(x^3 + 8) = 0 \]Set each factor equal to zero:\[ x = 0 \quad \text{or} \quad x^3 + 8 = 0 \]Solve \(x^3 + 8 = 0\) by factoring or using the cubic root:\[ x^3 = -8 \]\[ x = -2 \]The critical points are \(x = 0\) and \(x = -2\).

Step 3: Determine Local Maximum and Minimum

To identify the nature of the critical points, use the First Derivative Test.Evaluate the sign change of \(y'\):- Test intervals around \(x = 0\): Choose test points \(-1\) and \(1\).- For \(x = -1\), \(y'(-1) = 4(-1)^4 + 32(-1) = 4 - 32 = -28\) (negative).- For \(x = 1\), \(y'(1) = 4(1)^4 + 32(1) = 4 + 32 = 36\) (positive).Conclusion: \(x = -2\) is a local minimum.For \(x = 0\), the test suggests no sign change (from negative to positive), indicating a local minimum as well. Evaluate \(y\) function at these points to get y-values: \[ y(0) = -25, \quad y(-2) = \frac{4}{5}(-32) + 64 - 25 = -32 + 64 - 25 = 7\]

Step 4: Find the Second Derivative

Compute the second derivative to find the inflection points:Differentiate \(y'\):\[ y'' = \frac{d}{dx}(4x^4 + 32x) = 16x^3 + 32 \]

Step 5: Determine Inflection Points

Set the second derivative equal to zero to find potential inflection points:\[ 16x^3 + 32 = 0 \]Solve for \(x\):\[ 16x^3 = -32 \]\[ x^3 = -2 \]The real root of this equation is \(x = -\sqrt[3]{2}\), which needs to be checked in context; however, approximations suggest there's no visible change in concavity near this point for practical limits.

Step 6: Analyze Graph Relation with Derivatives

Graph \(y\), \(y'\), and \(y''\) together:- The points where the first derivative \(y'\) intersects the \(x\)-axis correspond to critical points (local maxima or minima).- The points where the second derivative \(y''\) intersects the \(x\)-axis suggest inflection points, though none exactly found in practical terms.The derivative graph's intersection at certain points can depict changes in the nature of the graph—like increasing or decreasing, or concave up or down transitions.

Key Concepts

First Derivative

The first derivative of a function provides crucial insights into the function's behavior. By differentiating the original function, we get the rate of change of the function, effectively revealing the slopes of the tangent lines at any point on the curve. For our function:

• Given: \[ y = \frac{4}{5} x^5 + 16x^2 - 25 \]
• The first derivative is found by differentiating with respect to \(x\):\[ y' = \left(4x^4 + 32x\right) \]

The first derivative helps us identify critical points, where the derivative is zero or undefined. These are potential locations for local maxima or minima on the graph. In practical terms, the first derivative graph intersects the \(x\)-axis at these critical points, offering a visual cue about potential extremas.

Second Derivative

The second derivative talks about the function's concavity—whether it's curving upwards or downwards at a certain point. By finding the second derivative, we can determine parts of the graph where the curvature changes. For the given function's first derivative:

• First derivative: \[ y' = 4x^4 + 32x \]
• Second derivative calculated as:\[ y'' = 16x^3 + 32 \]

Setting the second derivative to zero helps find potential inflection points. An inflection point is where the graph changes concavity from up to down or vice versa. In the step-by-step solution, the calculation aims for inflection by solving:\[ 16x^3 + 32 = 0 \]However, approximations indicate there might be no visible inflection within practical boundaries for this function. Thus, always check the concavity change to ensure a real inflection point.

Critical Points

Critical points are where the first derivative equals zero or is undefined, hinting at possible local extrema, either maxima or minima. For the given function, calculating the critical points involves:

• Setting the derivative \[ y' = 4x^4 + 32x \] to zero.
• Solve for \(x\) by factoring:\[ 4x(x^3 + 8) = 0 \]
• This gives critical points at \(x = 0\) and \(x = -2\).

To determine whether these critical points are maxima or minima, use a sign analysis of the derivative around these points. Often called the First Derivative Test, it involves testing the sign of \(y'\) just before and after the critical points:- For this function, tests indicate \(x = -2\) and \(x = 0\) as local minima based on changes from negative to positive.

Local Extrema

Local extrema refer to local maxima or minima on a graph—points where the function reaches a peak or dip locally. Identifying these points provides a deeper understanding of the function's shape and behavior across different intervals.

• For maxima, the function increases to the point and starts decreasing after.
• For minima, the function decreases to the point and starts increasing afterwards.

Using the First Derivative Test, the function's critical points at \(x = -2\) and \(x = 0\) were identified as potential local minima. Checking the function values at these critical points:

• At \(x = 0\), \(y(0) = -25\).
• At \(x = -2\), \(y(-2) = 7\).

This demonstrates that the graph effectively dips at \(x = 0\) and at \(x = -2\), marking out these local minima. Hence, inspecting the first derivative and evaluating \(y\) at critical points provide essential insight into the local behavior of the function.