Problem 34 - Applied Calculus

Question

If \(f(x)=(3 x+8)(2 x-5)\), find \(f^{\prime}(x)\) and \(f^{\prime \prime}(x)\).

Step-by-Step Solution

Answer

The first derivative is \( f'(x) = 12x + 1 \), and the second derivative is \( f''(x) = 12 \).

Step 1: Expand the Function

First, we need to expand the given function. The function is \( f(x) = (3x + 8)(2x - 5) \). Use the distributive property (or FOIL method) to expand this expression. Multiply each term in the first parenthesis by each term in the second parenthesis:\[(3x + 8)(2x - 5) = 3x \cdot 2x + 3x \cdot (-5) + 8 \cdot 2x + 8 \cdot (-5)\]Simplify this to:\[f(x) = 6x^2 - 15x + 16x - 40\]Combine like terms:\[f(x) = 6x^2 + x - 40\]

Step 2: Find the First Derivative

To find the first derivative \( f'(x) \), apply the power rule to each term of the polynomial \( f(x) = 6x^2 + x - 40 \). The power rule states \( \frac{d}{dx} [x^n] = nx^{n-1} \).The derivative of \( 6x^2 \) is \( 12x \), the derivative of \( x \) is \( 1 \), and the derivative of a constant \( -40 \) is \( 0 \).Therefore, the first derivative is:\[f'(x) = 12x + 1\]

Step 3: Find the Second Derivative

The second derivative \( f''(x) \) is the derivative of the first derivative \( f'(x) \). From the previous step, we found \( f'(x) = 12x + 1 \).Apply the power rule again. The derivative of \( 12x \) is \( 12 \), and the derivative of a constant \( 1 \) is \( 0 \).Therefore, the second derivative is:\[f''(x) = 12\]

Key Concepts

Derivatives

A derivative in calculus is a fundamental concept that measures how a function changes as its input changes. Imagine you're driving a car, and the speedometer shows how fast you're going—that's like what a derivative does for a function.
When we talk about finding the derivative of a function, we're looking for its "rate of change." In the context of this exercise, if we have a function representing a curve, its derivative gives us the slope of the tangent line at any point on that curve.
For a function like the one in the exercise, behavior can be observed around particular points using derivatives. For instance:

• The first derivative tells us if the function is increasing or decreasing.
• The second derivative gives information about the curvature of the graph, helping us understand concepts like concavity and inflection points.

Overall, derivatives are powerful because they help analyze and predict trends in simple to complex functions.

Power Rule

The power rule is an efficient shortcut used in differentiation, which is part of calculus. It allows us to quickly compute derivatives for polynomial expressions.
When you apply the power rule, it involves the simple operation of taking a power down as a coefficient and then reducing the power by one.
Here's how the rule works:

• If you have a term like \(x^n\), its derivative is \(nx^{n-1}\).

To illustrate:

• The derivative of \(x^3\) is \(3x^2\) because you multiply by the exponent 3 and then reduce the exponent by 1.
• Similarly, for \(6x^2\), the power rule gives a derivative of \(12x\) by multiplying by 2 and lowering the power by one.

This rule provides an efficient way to handle derivatives of polynomials, saving time and effort when dealing with multiple terms.

Polynomial Function Expansion

Expanding polynomial functions is essential in calculus, especially before applying derivative rules. In this exercise, expansion was the initial step to simplifying the compound expression.
When dealing with functions like \((3x + 8)(2x - 5)\), expansion involves using the distributive property or the FOIL method to remove the parentheses and combine like terms.
Steps in expansion:

• Distribute each term in the first polynomial across every term in the second.
• Simplify the resulting expressions by adding or subtracting like terms (those with the same power of \(x\)).

For example, when expanding \((3x + 8)(2x - 5)\), you'd calculate:

• \(3x \cdot 2x = 6x^2\)
• \(3x \cdot (-5) = -15x\)
• \(8 \cdot 2x = 16x\)
• \(8 \cdot (-5) = -40\)

Combine to get \(6x^2 + x - 40\), a polynomial that's now ready for differentiation. This preparation step simplifies working with derivatives, allowing easy application of rules like the power rule.