Problem 8 - 5 Steps to A 5: AP Calculus AB 2021

Question

If \(f^{\prime \prime}(x)=x^{2}(x+3)(x-5),\) find the values of \(x\) at which the graph of \(f\) has a change of concavity.

Step-by-Step Solution

Answer

Answer: The graph of the function f has a change of concavity at x = -3, x = 0, and x = 5.

Step 1: Write down the second derivative

Given the second derivative of the function: $$f^{\prime\prime}(x) = x^{2}(x+3)(x-5)$$

Step 2: Find critical points of the second derivative

To find the critical points, we need to find the points where the second derivative is zero or undefined. $$f^{\prime\prime}(x) = x^2 (x + 3)(x - 5) = 0$$ This equation is satisfied when x = -3, 0, or 5.

Step 3: Analyze sign changes to determine intervals

Now let's determine the intervals where the second derivative is positive or negative: - For \(x<-3\), all three factors \(x^2\), \((x+3)\) and \((x-5)\) are negative. The product of three negative numbers is negative, so \(f^{\prime\prime}(x) < 0\) in this interval. - For \(-3 0\) in this interval. - For \(x>5\), all three factors \(x^2\), \((x+3)\) and \((x-5)\) are positive. The product of three positive numbers is positive, so \(f^{\prime\prime}(x) > 0\) in this interval.

Step 4: Identify the change of concavity

The graph of \(f\) has a change of concavity when \(f^{\prime\prime}(x)\) changes its sign. We can see that the second derivative changes its sign exactly at the points x = -3, 0, and 5. So, the graph of \(f\) has a change of concavity at \(x=-3\), \(x=0\), and \(x=5\).

Key Concepts

Second Derivative Test

The Second Derivative Test is a handy tool in calculus for determining whether a critical point is a local maximum, local minimum, or a saddle point (neither). To perform this test, you first need to find the first derivative of the function and identify the critical points where this derivative is zero or undefined.

After locating these critical points, you evaluate the second derivative at these points. If the second derivative is positive at a critical point, the function has a local minimum there; if it's negative, there's a local maximum. However, if the second derivative is zero, the test is inconclusive - this is where inflection points might come into play, which are closely related but not the same thing as turning points established by this test.

Application of the Second Derivative Test

In our exercise example, the function given is already the second derivative of an unprovided original function, implying we've bypassed finding critical points with the first derivative. Instead, we directly apply the second derivative test to find where concavity changes, which can signal potential inflection points.

Critical Points

Critical points in calculus are where the first derivative of a function is either zero or does not exist. At these points, the function can have a local maximum or minimum. Moreover, the behavior of a function can change drastically at these points, making their determination vital for understanding the function's graph.

To find critical points, you first calculate the first derivative of the function and then solve for when this derivative equals zero or is undefined. However, critical points are not guaranteed to be maxima or minima; the Second Derivative Test can provide further insights into the nature of these points.

Finding Critical Points from the Second Derivative

While critical points typically come from the first derivative, in our problem, we find values where the second derivative equals zero, which are analogous to 'second-order' critical points. These are the x-values where changes in concavity may occur and can suggest the presence of an inflection point.

Inflection Points

Inflection points occur where a function's graph changes concavity, from concave up to concave down, or vice versa. These points are essential in understanding the geometry of the function's graph, as they indicate significant changes in the behavior of the curve.

To potentially locate inflection points, we look for values where the second derivative changes sign. A change in sign indicates a change in concavity and thus a potential inflection point. However, not every zero of the second derivative guarantees an inflection point; the sign must change.

Determining Inflection Points from the Exercise

In our example, after finding the x-values where the second derivative equals zero (x = -3, 0, 5), we examine the intervals between these values to check for sign changes in the second derivative. The changes in the sign of the second derivative, from negative to positive or positive to negative, confirm that at x = -3, 0, and 5, our function experiences a change in concavity, thus identifying these points as inflection points.