Problem 37
Question
In Exercises, find the second derivative and solve the equation \(f^{\prime \prime}(x)=0\). $$ f(x)=(x+3)(x-4)(x+5) $$
Step-by-Step Solution
Verified Answer
The second derivative \(f''(x) = 6x - 2\), and \(x = 1/3\) is the solution for \(f''(x) = 0\).
1Step 1: 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 \(f'(x) = 3x^2 - 2x -15\)
2Step 2: 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 \(nx^{n-1}\)
3Step 3: 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 RuleMastering the Power RuleSolving Calculus Problems Efficiently
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:
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.
The product rule is formulated as follows:
- If you have a product of two functions, say \(g(x)\) and \(h(x)\), their derivative is \(g'(x)h(x) + g(x)h'(x)\).
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: \(f'(x) = 3x^2 - 2x - 15\).
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.
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.
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 \(f''(x) = 0\).
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.
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 \(x = \frac{1}{3}\).
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.
Other exercises in this chapter
Problem 37
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