Problem 117 - Thomas Calculus

Question

Suppose that the second derivative of the function \(y=f(x)\) is \begin{equation}y^{\prime \prime}=(x+1)(x-2).\end{equation} For what \(x\) -values does the graph of \(f\) have an inflection point?

Step-by-Step Solution

Answer

The graph of \(f\) has inflection points at \(x = -1\) and \(x = 2\).

Step 1: Understand the Inflection Point Criterion

An inflection point occurs when the second derivative changes sign. We need to determine where \( y''=(x+1)(x-2) \) changes sign.

Step 2: Find Critical Points of the Second Derivative

The critical points of \( y'' \) occur where \((x+1)(x-2) = 0\). Set each factor equal to zero and solve separately: \(x+1=0\) and \(x-2=0\). This gives \(x=-1\) and \(x=2\).

Step 3: Test Intervals Between Critical Points

Choose test points from the intervals: \((-\infty, -1)\), \((-1, 2)\), and \((2, \infty)\).- For \(x < -1\), choose \(x = -2\): \(y''=(-2+1)(-2-2)=(-1)(-4)=4 > 0\).- For \(-1 < x < 2\), choose \(x = 0\): \(y''=(0+1)(0-2)=1(-2)=-2 < 0\).- For \(x > 2\), choose \(x = 3\): \(y''=(3+1)(3-2)=4(1)=4 > 0\).

Step 4: Identify Sign Change and Inflection Points

The sign of \(y''\) changes from positive to negative at \(x = -1\) and from negative to positive at \(x = 2\). Therefore, the function \(f(x)\) has inflection points at \(x = -1\) and \(x = 2\).

Key Concepts

Second Derivative

The second derivative of a function, denoted as \( y'' \), is a vital tool in determining the concavity of the function's graph. When you have \( y'' = (x+1)(x-2) \), you're looking at how the slope of the original function \( f(x) \) is changing. The second derivative tells us whether the graph of \( f(x) \) is curving upwards or downwards.
- When \( y'' > 0 \), the graph is concave up. - When \( y'' < 0 \), the graph is concave down.
An inflection point occurs where this concavity changes, meaning \( y'' \) changes its sign. Understanding the second derivative is key in calculus as it helps us identify points where the graph's shape alters.

Critical Points

Critical points related to the second derivative are where the function's concavity might switch. These are the \( x \)-values where \( y'' = 0 \) or doesn't exist. For our specific problem, the equation \( (x+1)(x-2) = 0 \) provides us with the critical points \( x = -1 \) and \( x = 2 \).
These values are candidates for inflection points. But it's necessary to test the intervals around these points to confirm if a sign change actually occurs. Finding these critical points allows us to focus our attention on where possible curvature changes might be happening on the graph.

Sign Change

The concept of a sign change in the second derivative is integral in identifying inflection points. Once you have the critical points, the next step is to investigate if \( y'' \) changes its sign across these points.
For example, by testing values in intervals determined by \(-1\) and \(2\), you will check the sign of \( y'' \) in each interval:

• Choose \( x = -2 \) for the interval \((-\infty, -1)\); \( y'' > 0 \).
• Choose \( x = 0 \) for the interval \((-1, 2)\); \( y'' < 0 \).
• Choose \( x = 3 \) for \((2, \infty)\); \( y'' > 0 \).

In this analysis, \( y'' \) switches from positive to negative at \( x = -1 \) and negative to positive at \( x = 2 \). This confirms inflection points exist at these values.

Interval Testing

Interval testing is a method that uses selected values to determine the behavior of a function's second derivative across different sections of the real number line. This approach ensures that sign changes do occur at critical points, validating them as true inflection points.
By selecting test points in the intervals created by your critical values, you can examine the nature of \( y'' \) in each segment:
- In \((-\infty, -1]\), choose a test point like \( x = -2 \). Here, \( y'' \) is positive.- Between \(-1\) and \(2\), use \( x = 0 \). Here, \( y'' \) becomes negative.- Beyond \(2\), with \( x = 3 \), \( y'' \) returns to being positive.
These test results, showing an alternating pattern in the sign of \( y'' \), are what ultimately confirm the presence of inflection points at \( x = -1 \) and \( x = 2 \). Thus, interval testing is a strategic method in verifying where these crucial points occur.