Problem 1 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Verify directly that \(F\) is an antiderivative of \(f\) $$F(x)=\frac{1}{3} x^{3}+2 x^{2}-x+2 ; f(x)=x^{2}+4 x-1$$

Step-by-Step Solution

Answer

To verify that \(F(x) = \frac{1}{3}x^3 + 2x^2 - x + 2\) is an antiderivative of \(f(x) = x^2 + 4x - 1\), we calculate the derivative of F(x) directly using the power rule. Doing so, we find \(F'(x) = x^2 + 4x - 1\). Since \(F'(x) = f(x)\), we have verified that F is an antiderivative of f.

Step 1: Identify the functions

First, let's identify the given functions. We are given: \[F(x) = \frac{1}{3}x^3 + 2x^2 - x + 2\] \[f(x) = x^2 + 4x - 1\] Our goal is to show that F'(x) = f(x).

Step 2: Calculate the derivative of F(x)

Now, let's find the derivative of F(x). Apply the power rule of the derivative, which states that if F(x) = ax^n, where a and n are constants, then its derivative F'(x) = nax^(n-1). \[F'(x) = (\frac{1}{3} * 3)x^{3-1} + (2*2)x^{2-1} - 1x^{1-1} + 0\]

Step 3: Simplify the derivative of F(x)

After applying the power rule, simplify the expression for F'(x): \[F'(x) = x^2 + 4x - 1\]

Step 4: Compare F'(x) and f(x)

We have calculated the derivative F'(x) and found that: \[F'(x) = x^2 + 4x - 1\] This is exactly the function f(x) that we were given: \[f(x) = x^2 + 4x -1\] Since F'(x) = f(x), we have now directly verified that F is an antiderivative of f.

Key Concepts

Power Rule

The power rule is a quick way to find the derivative of a function in the form of a power of x. Specifically, if you have a term like \( ax^n \), where \( a \) and \( n \) are constants, the power rule helps you determine its derivative without much fuss. Here's the secret in a nutshell: bring the power down and reduce it by one.

For example, using the power rule, the derivative of \( ax^n \) becomes \( nax^{n-1} \). It's about taking the exponent (\( n \)), multiplying it by the coefficient (\( a \)), and then decreasing the exponent by 1. Simple, right?

In our case, each term of \( F(x) = \frac{1}{3} x^3 + 2x^2 - x + 2 \) uses the power rule to find the derivative:

• \( \frac{1}{3} \times 3x^{3-1} = x^2 \)
• \( 2 \times 2x^{2-1} = 4x \)
• The term \( -x \) simplifies to \( -1 \)
• The constant \( 2 \) disappears as it contributes nothing to the derivative

The power rule makes finding derivatives easy peasy.

Derivative

A derivative represents the rate at which a function is changing at any given point. Think of it as the 'instantaneous speed' of the function. When we calculate a derivative, we're essentially finding that speed.

In calculus, the process to differentiate a function involves applying specific rules, like the power rule, to find the derivative. Calculating the derivative of \( F(x) \) gives us \( F'(x) \), the slope of the tangent line to the function at any point.

In the example, we found that the derivative of \( F(x) = \frac{1}{3} x^3 + 2x^2 - x + 2 \) is \( F'(x) = x^2 + 4x - 1 \). Every component of the function modifies how F is tilting or changing its path.

• \( x^2 \) contributes a parabolic curve
• A linear element like \( 4x \) shifts the slope
• The constant \( -1 \) adjusts the baseline value

These combined give us the rate of change accurately reflected by \( F'(x) \).

Function Verification

Verifying that one function is the antiderivative of another essentially checks if differentiation can be reversed. If differentiating \( F(x) \) gives us the function \( f(x) \), then \( F(x) \) is the antiderivative of \( f(x) \).

To verify, we compare the derivative \( F'(x) \) and the given function \( f(x) \). Suppose we learned that \( F'(x) = x^2 + 4x - 1 \). Also, we know \( f(x) = x^2 + 4x - 1 \). They match perfectly!

This is verification in practice:

• Calculate \( F'(x) \) using established rules, like the power rule
• Compare \( F'(x) \) to \( f(x) \)
• If they are identical, the antiderivative claim holds true

This straightforward process allows us to check our work and confirms that after finding the derivative, we have circled back to the original function \( f(x) \). Such steps ensure that the initial function describes accurate changes when reviewed through its antiderivative.