Problem 109 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understand the Inflection Point Condition

An inflection point occurs where the second derivative, \(y''\), changes its sign. This means that the graph of the function changes concavity.

Step 2: Set the Second Derivative to Zero

To find potential inflection points, we first solve the equation \((x+1)(x-2) = 0\). This gives us the critical points where the concavity might change.

Step 3: Solve the Equation

Solve \((x+1)(x-2) = 0\) to find \(x\)-values: \(x+1=0\) yields \(x=-1\), and \(x-2=0\) yields \(x=2\). These are the points where the second derivative might change signs.

Step 4: Test for Sign Changes

Examine the intervals around \(x=-1\) and \(x=2\). Consider the intervals: \((-\infty, -1)\), \((-1, 2)\), and \((2, \infty)\). Substitute a test value from each interval into the second derivative to determine concavity signs.

Step 5: Determine Concavity on Intervals

Calculate \(y''\) for a value in each interval:- For \(x < -1\) (e.g., \(x=-2\)), \((x+1)(x-2) = (0)(-4) > 0\).- For \(-1 < x < 2\) (e.g., \(x=0\)), \((x+1)(x-2) = (1)(-2) < 0\).- For \(x > 2\) (e.g., \(x=3\)), \((x+1)(x-2) = (4)(1) > 0\).

Step 6: Identify Location of Inflection Points

From the sign changes:- From positive to negative at \(x=-1\) (concavity changes from up to down).- From negative to positive at \(x=2\) (concavity changes from down to up).Thus, inflection points are at \(x=-1\) and \(x=2\).

Key Concepts

Second Derivative Test

The second derivative test is a helpful tool to identify points of interest on a graph, particularly regarding the concavity of a function. If you have the function's second derivative, this test helps determine whether a point is a local maximum, a local minimum, or an inflection point. To apply this test:

• Compute the second derivative of the function, denoted as \(f''(x)\).
• Find critical points by setting the first derivative, \(f'(x)\), to zero and solving for \(x\).
• Apply the second derivative to these critical points.
• A positive \(f''(x)\) at a critical point suggests a local minimum, while a negative \(f''(x)\) indicates a local maximum.
• If \(f''(x)\) equals zero or changes sign, further investigation is necessary, as this may indicate an inflection point.

If the sign of the second derivative changes as you move through a critical point, you're likely observing an inflection point. The implications of applying this test can help in understanding how the graph behaves, notably where it curves up or down.

Concavity

Concavity describes how a graph curves. A graph is concave up if it looks like a cup and concave down if it looks like a cap. Mathematically, the concavity of a function is determined based on its second derivative. Here's how it works:

• If \(f''(x) > 0\) for an interval, the graph is concave up on that interval.
• If \(f''(x) < 0\) for an interval, the graph is concave down on that interval.

To understand this better, consider an example:

• When a graph is concave up, the slope of the tangent line increases as \(x\) increases; graphs for positive intervals of second derivatives resemble a smile.
• Conversely, for a concave down graph, the tangent line slope decreases as \(x\) increases. These graphs resemble a frown.

The concavity helps to visualize how a function behaves beyond simply increasing or decreasing, offering insights into the overall shape and potential changes in direction.

Critical Points

Critical points are vital in calculus as they indicate where a function's graph has horizontal tangents. To locate these points, you take the first derivative of a function and set it to zero, then solve for \(x\). Here's more on this topic:

• Critical points occur where \(f'(x) = 0\) or where \(f'(x)\) is undefined.
• At these points, the function might have a local maximum, minimum, or a point of inflection.

Why are critical points so significant?

• They help in understanding the function's overall behavior, guiding us to local peaks, troughs, and potential inflection points.
• By examining what happens at these points and applying further tests (like the second derivative test), we can predict and sketch the function's graph accurately.

Remember, while critical points provide major clues about the function's behavior, it's important to analyze further to determine their role — as a peak, trough, or a point of curvature change.