Question

In Exercises, find the second derivative and solve the equation \(f^{\prime \prime}(x)=0\). $$ f(x)=(x+3)(x-4)(x+5) $$

Short answer

The second derivative \(f''(x) = 6x - 2\), and \(x = 1/3\) is the solution for \(f''(x) = 0\).

Step-by-step solution

Find the first derivative

The first derivative of \(f(x)\) can be obtained by applying the product rule. The product rule states that the derivative of the product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function. Given \(f(x) = (x+3)(x-4)(x+5)\), let’s apply the product rule sequentially. Take \(g(x) = (x+3)\) and \(h(x) = (x-4)(x+5)\), then \(f'(x) = g'(x)h(x) + g(x)h'(x)\), which simplifies to \(f'(x) = (1)(x-4)(x+5) + (x+3)(x-1)\), which further simplifies to \(f'(x) = 3x^2 - 2x -15\)

Find the second derivative

To find the second derivative, we differentiate f'(x). Since \(f'(x) = 3x^2 - 2x -15\), \(f''(x) = 6x - 2\) by applying the power rule, which states that the derivative of \(x^n\) is \(nx^{n-1}\)

Solve the second derivative equation

For \(f''(x)=0\), this implies that \(6x - 2 = 0\). Solving for \(x\), we find that \(x = 2/6 = 1/3.\) This is the point where the graph of the function has a zero slope or a 'tangent' horizontal line

Key concepts

Understanding the Product Rule

When faced with a function that is the product of multiple terms, like \((x+3)(x-4)(x+5)\), the product rule in calculus becomes essential. This rule helps us differentiate the product of two or more functions.
The product rule is formulated as follows:

• If you have a product of two functions, say \(g(x)\) and \(h(x)\), their derivative is \(g'(x)h(x) + g(x)h'(x)\).

In our original problem, we applied it in a sequence due to the three terms involved.
First, separate the functions: \(g(x) = (x+3)\) and \(h(x) = (x-4)(x+5)\). By finding each function's derivative and applying the rule, we find \(f'(x) = (1)(x-4)(x+5) + (x+3)(x-1)\).
We combined like terms and simplified to get the derivative \(f'(x) = 3x^2 - 2x -15\).
This process demonstrates the utility of the product rule in breaking down complex expressions into manageable pieces and is a fundamental technique often used in calculus problems.

Mastering the Power Rule

The power rule is one of the simplest derivative rules but it's also incredibly powerful. Anytime you see a term raised to a power, this rule is your best friend.
The power rule states that for any function \(x^n\), its derivative is \(nx^{n-1}\).
In our initial problem, after applying the product rule, we arrived at the first derivative: \(f'(x) = 3x^2 - 2x - 15\).

• To find \(f''(x)\), we differentiated each term individually using the power rule.

- For \(3x^2\), we used the power rule to get \(6x\).- For \(-2x\), the power rule gives us \(-2\). - As \(-15\) does not contain \(x\), it becomes \(0\).
Thus, the second derivative is \(f''(x) = 6x - 2\). Recognizing when and how to apply the power rule allows us to effectively handle calculus problems, particularly when dealing with polynomials.

Solving Calculus Problems Efficiently

Calculus problems often require interpreting and solving derivatives, like finding when a function's derivative is zero. This is crucial for determining points where the slope of a tangent line is horizontal.
In this task, after finding the second derivative \(f''(x) = 6x - 2\), we solved the equation \(f''(x) = 0\).

• Setting the equation to zero requires isolating \(x\): \(6x - 2 = 0\).
• Solving for \(x\) gives \(x = \frac{1}{3}\).

By solving for \(x\), we found the critical point where the rate of change of the slope is zero.
This means at \(x = \frac{1}{3}\), the graph of the original function has a tangent that is completely horizontal.
Understanding how to solve such problems gives insight into the behavior of functions and is a key aspect of calculus: noticing changes and predicting trends through derivatives.