Problem 38 - Calculus an Applied Approach

Question

find the second derivative and solve the equation \(f^{\prime \prime}(x)=0\) $$ f(x)=(x+2)(x-2)(x+3)(x-3) $$

Step-by-Step Solution

Answer

The inflection point of the function \(f(x)= (x+2)(x-2)(x+3)(x-3)\) occurs at \(x = 0\).

Step 1: Apply the First Derivative using Product Rule

To apply the product rule of derivative, consider function \(f(x)\) as a product of two functions namely \(h(x) = (x+2)(x-2) = x^2-4\) and \(g(x) = (x+3)(x-3) = x^2-9\). The first derivative of \(f(x)\) which is \(f'(x)\) is computed using the product rule as follows: \(f'(x) = h'(x)g(x) + h(x)g'(x)\). By simple differentiation, we have \(h'(x) = 2x\), and \(g'(x)=2x\), thus, \(f'(x) = 2x(x^2-9) + 2x(x^2-4) = 4x(x^2-4+9-4) = 4x^3\).

Step 2: Apply the Second Derivative

Next, apply the power rule which is simplified derivative to get the second derivative of \(f(x)\). The second derivative of \(f(x)\), denoted as \(f''(x)\), is the derivative of \(f'(x)=4x^3\), thus \(f''(x)=12x^2\).

Step 3: Equate Second Derivative to Zero and Solve

Lastly, set \(f''(x)\) equal to zero and solve for \(x\). This gives the inflection points of the original function \(f(x)\). \[12x^2=0\]\[-1.5ex] Solving this results in \(x=0\).

Key Concepts

Second Derivative

The concept of a second derivative is crucial in understanding how the rate of change of a function's slope behaves. In calculus, while the first derivative tells us about the slope of the tangent line at any point on a function, the second derivative gives us information about the curvature. More specifically, it indicates whether the function is concave up or concave down at a particular interval.
Understanding the process:

• If the second derivative, denoted as \( f''(x) \), is positive, the function \( f(x) \) is concave up. Think of it like a cup which can "hold water".
• If \( f''(x) \) is negative, the function is concave down, similar to a frown.
• When \( f''(x) = 0 \), it indicates potential inflection points where the function might change its concavity.

This insight into a function's behavior is invaluable in curve sketching and optimizing problems. In the provided exercise, finding the second derivative \( f''(x) = 12x^2 \) and equating it to zero helps locate the point \( x = 0 \), a crucial inflection point for the function \( f(x) \).

Product Rule

The product rule is fundamental when you need to take the derivative of a product of two functions. Imagine you have two differentiable functions, say \( u(x) \) and \( v(x) \). The product rule states that the derivative of their product is given by:
\[ (uv)' = u'v + uv' \]
Here's a step-by-step breakdown:

• Differentiate one function while keeping the other constant.
• Then, sum the result with the differentiation of the second function while keeping the first function constant.

This rule is not just a mechanical tool but helps grasp how two changing quantities affect each other's rate of change. In our exercise, we applied the product rule on \( h(x) = (x+2)(x-2) \) and \( g(x) = (x+3)(x-3) \). By finding \( h'(x) \) and \( g'(x) \), substituting them back into the product rule formula allowed us to find the first derivative \( f'(x) = 4x^3 \), leading towards determining the second derivative.

Solving Equations

Solving equations, particularly in calculus, often involves finding the values of \( x \) that satisfy a given mathematical expression. When we explore equations of the type \( f''(x) = 0 \), we're on the hunt for important features of the function, such as inflection points.
Here is a simple approach:

• First, compute the necessary derivative based on the equation type; in our scenario, it is the second derivative \( f''(x) \).
• Set the result to zero, forming a new, more straightforward equation.
• Solve for \( x \) to find the points of interest.

In this particular exercise, solving \( 12x^2 = 0 \), we quickly find \( x = 0 \). This point is significant because it marks where the function \( f(x) \) may switch concavity. Hence, addressing these points can reveal a lot about the behavior of the original function.