Problem 47 - Calculus Single Variable

Question

The second derivative \(f^{\prime \prime}\) of a function \(f\) is given. Determine every \(x\) at which \(f\) has a point of inflection. $$ f^{\prime \prime}(x)=\left(x^{2}-25\right)^{3} $$

Step-by-Step Solution

Answer

The function has inflection points at \(x = 5\) and \(x = -5\).

Step 1: Understand the Condition for Inflection Points

A point of inflection occurs where the second derivative changes sign. This means that if we have a function's second derivative, \(f''(x)\), an inflection point will occur at values of \(x\) where \(f''(x) = 0\) and \(f''(x)\) changes sign around these points.

Step 2: Set the Second Derivative to Zero

Find the values of \(x\) where the second derivative is zero. For this exercise:\[ f''(x) = (x^2 - 25)^3 = 0 \]Solve this equation for \(x\). Since \((x^2 - 25)^3 = 0\), we first find \(x^2 - 25 = 0\).

Step 3: Solve for \(x\)

From \(x^2 - 25 = 0\), solve for \(x\):\[x^2 = 25\]This gives:\[x = \pm 5\]Thus, the potential points of inflection are at \(x = 5\) and \(x = -5\).

Step 4: Verify Sign Change

Ensure that the second derivative changes sign around these solutions. Test intervals around \(x = 5\) and \(x = -5\) to determine if the sign of \(f''(x)\) changes:- Choose points like 0 (between \(x = -5\) and \(x = 5\)), 6 (just past \(x = 5\)), and -6 (just before \(x = -5\)).- Note that the cube in \(g(x) = (x^2 - 25)^3\) implies that the sign changes when \(x\) crosses past the values of 5 and -5.

Step 5: Conclusion

Since \(f''(x)\) changes signs at both \(x = 5\) and \(x = -5\), these are points of inflection.

Key Concepts

Second Derivative

The second derivative of a function, denoted as \( f''(x) \), provides valuable insights into the curvature or concavity of the function. Essentially, it informs us how the slope of the tangent line to the curve is changing. When analyzing a graph:

• If \( f''(x) > 0 \), the graph is concave up, similar to a smiling face.
• If \( f''(x) < 0 \), the graph is concave down, resembling a sad face.
• The points where the concavity changes are potential inflection points, making the understanding of \( f''(x) \) critical.

These characteristics help in identifying where the behavior of the function shifts, providing richer details of the function's shape and flow.

Sign Change

To determine inflection points, we must look for sign changes in the second derivative. A sign change means that the derivative went from positive to negative, or vice versa.
Consider the function \( f''(x) = (x^2 - 25)^3 \). The key is to find where \( f''(x) = 0 \), i.e., where a potential sign change could occur.

• Identify values of \( x \) where the derivative is nullified: \( (x^2 - 25)^3 = 0 \).
• This equation simplifies to \( x^2 - 25 = 0 \), giving \( x = \pm 5 \).

For a valid inflection point, \( f''(x) \) must alter its sign as you pass through these \( x \)-values. Hence, testing values on either side of \( x = 5 \) and \( x = -5 \) will clarify if a sign change occurs.

Solve Equations

The process of solving equations plays a pivotal role when identifying inflection points. To find where \( f''(x) = 0 \), you solve the equation for \( x \). Let's break it down:

• Given the equation \( (x^2 - 25)^3 = 0 \), we take the cube root to isolate \( x^2 - 25 \).
• This yields \( x^2 - 25 = 0 \), removing the exponent of 3 simplifies our task.
• Proceed to solve \( x^2 = 25 \) to find \( x = \pm 5 \).

Through these algebraic steps, we've uncovered the locations where potential sign changes—and thus inflection points—might exist.

Critical Points

Critical points of a function are where its derivatives are zero or undefined. In the context of inflection points, we're particularly interested in where the second derivative equals zero.
In our exercise, we determined critical points to be \( x = 5 \) and \( x = -5 \). However, unlike critical points where the first derivative equals zero, inflection points also require a sign change in the second derivative.

• Test intervals around these critical points (like \( x = -6, 0, 6 \)) to verify if a sign change occurs in \( f''(x) \).
• Checking values between and just outside your critical points will ensure any oscillation between positive and negative values is detected.

By confirming these tests, you validate \( x = 5 \) and \( x = -5 \) as points of inflection, offering a clear picture of how the curve behaves at these key locations.