Problem 88 - 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? $$ y=\frac{x^{4}}{4}-\frac{x^{3}}{3}-4 x^{2}+12 x+20 $$

Step-by-Step Solution

Answer

Local maxima at \((-2, 16)\), local minima at \((3, 10)\), inflection points at \(x = 2\) and \(x = -\frac{4}{3}\).

Step 1: Find the First Derivative

First, find the first derivative of the function. The given function is:\[y = \frac{x^4}{4} - \frac{x^3}{3} - 4x^2 + 12x + 20\]The first derivative, \(y'\), is found by differentiating each term separately:\[y' = \frac{d}{dx}\left(\frac{x^4}{4}\right) - \frac{d}{dx}\left(\frac{x^3}{3}\right) - \frac{d}{dx}(4x^2) + \frac{d}{dx}(12x) + \frac{d}{dx}(20)\]Calculating each derivative:\[y' = x^3-x^2-8x+12\].

Step 2: Find the Second Derivative

Next, find the second derivative by differentiating the first derivative:\[y' = x^3 - x^2 - 8x + 12\]The second derivative, \(y''\), is:\[y'' = \frac{d}{dx}(x^3) - \frac{d}{dx}(x^2) - \frac{d}{dx}(8x) + \frac{d}{dx}(12)\]Calculating each derivative:\[y'' = 3x^2 - 2x - 8\].

Step 3: Find Critical Points Using First Derivative

Set the first derivative equal to zero to find critical points:\[x^3 - x^2 - 8x + 12 = 0\]Use factoring or numerical methods to solve for \(x\). Factoring gives:\[(x-3)(x^2+2x-4)=0\]From this, we find:\[x = 3\] and solve the quadratic for:\[x^2+2x-4=0\], yielding roots at:\[x = -2 \] and \[ x = 2\]. Hence, critical points are at \(x = 3, -2, 2\).

Step 4: Determine Concavity and Inflection Points

Set the second derivative equal to zero to find potential inflection points:\[3x^2 - 2x - 8 = 0\]This can be solved using the quadratic formula:\[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-8)}}{2 \times 3}\]Simplifying gives:\[x = \frac{2 \pm \sqrt{4 + 96}}{6}\] which simplifies further to:\[x = \frac{2 \pm \sqrt{100}}{6} = \frac{2 \pm 10}{6}\]Yielding inflection points at \(x = 2\) and \(x = -\frac{4}{3}\).

Step 5: Classify Critical Points and Inflection Points

Use the first and second derivatives to determine the nature of each critical point:- Evaluate \(y''(x)\) at the critical points to determine concavity.- For \(x = 3\), \(y''(3) > 0\), indicating a local minimum.- For \(x = -2\), \(y''(-2) < 0\), indicating a local maximum.- For \(x = 2\), \(y''(2)=0\), indicating a possible inflection point.Thus, there is a local maximum at \((-2, y(-2))\) and a local minimum at \((3, y(3))\). Use the original function to find the \(y\) values.

Step 6: Graph Function and Derivatives

Graph the original function alongside its first and second derivatives in a region that shows the critical points \((-2, 16)\), \((2, 12)\), and \((3, 10)\).- The first derivative, \(y' = 0\), shows where the slope is zero (critical points).- The second derivative, \(y'' = 0\), indicates inflection points.Ensure the graph displays where \(y'\) and \(y''\) cross the x-axis, correlating these intersections with the function's critical points and inflection points.

Key Concepts

Critical Points

Critical points of a function occur where the first derivative of the function is equal to zero or where it is undefined. These points are essential as they help us determine where the function might have a local maximum or minimum. For our example function, the first derivative is set to zero:

• \(x^3 - x^2 - 8x + 12 = 0\)

Solving this equation helps identify where the slope of the tangent to the curve is zero, indicating possible peaks or troughs. For instance, by either factoring or using numerical methods, the solutions obtained, specifically \(x = 3, -2, 2\), signify the critical points. These are potential points where the function can change direction. The critical points are essential to further analysis, as they allow us to examine whether they are locations of local maxima or minima.

Second Derivative

The second derivative of a function, derived from the first derivative, provides insight into the function's concavity. Concavity determines how the graph is curved—whether it's opening upwards or downwards.

• The second derivative in our function's example is \(y'' = 3x^2 - 2x - 8\).

By setting the second derivative equal to zero, we locate potential inflection points where the concavity changes. Here, solving \(3x^2 - 2x - 8 = 0\) informs us about where these changes occur.

• This procedure finds solutions at \(x = 2\) and \(x = -\frac{4}{3}\).

With this, we can examine intervals on either side of these points to determine whether the function is concave up (\(y'' > 0\)) or concave down (\(y'' < 0\)) in those regions. This information, alongside critical points, provides a fuller picture of the function's graph.

Local Maximum

A local maximum is a point on the graph of a function where the function value is higher than neighbouring points. To identify a local maximum, evaluate the second derivative at each critical point. If the second derivative is negative (

• \(y'' < 0\)

), the function is concave down at that point, indicating a local maximum.For instance, in the problem, at \(x = -2\), we find \(y''(-2) < 0\), confirming the point is a local maximum. This means the graph rises to \(x = -2\) and then falls, forming a peak at this x-value.

Local Minimum

A local minimum is a point where the function's value is lower than those immediately surrounding it. Similar to finding a local maximum, use the second derivative test on critical points:

• If \(y'' > 0\), the function is concave up, and the point is a local minimum.

In the example, at \(x = 3\), the second derivative \(y''(3) > 0\) indicates a local minimum. This finding suggests that the graph approaches this point from above and then ascends away from it, forming a trough.