Problem 37 - Thomas Calculus in SI Units

Question

Find the function with the given derivative whose graph passes through the point \(P\). $$f^{\prime}(x)=2 x-1, \quad P(0,0)$$

Step-by-Step Solution

Answer

The function is \( f(x) = x^2 - x \).

Step 1: Integrate the derivative

To find the original function, we need to integrate the given derivative. The derivative of the function is \( f'(x) = 2x - 1 \). The integral of this function gives us the original function plus a constant of integration \( C \). This can be written as: \[ f(x) = \int (2x - 1)\, dx \]

Step 2: Perform the integration

Let's find the integral of \( 2x - 1 \). When we integrate, we use the power rule \( \int x^n dx = \frac{x^{n+1}}{n+1} \) for each term: \[ \int 2x\, dx = x^2 \] and \[ \int -1\, dx = -x \]. Combining these, the integral becomes: \[ f(x) = x^2 - x + C \]

Step 3: Solve for the constant using the given point

The function passes through the point \( P(0, 0) \). This means when \( x = 0 \), \( f(x) = 0 \). Substituting these into the function: \[ f(0) = 0^2 - 0 + C = 0 \]. Solving for \( C \), we find that \( C = 0 \).

Step 4: Write the final function

With \( C = 0 \), we can write the final function as: \[ f(x) = x^2 - x \]. This is the function whose derivative is \( 2x - 1 \) and which passes through the point \( P(0,0) \).

Key Concepts

Derivative

The derivative of a function is a fundamental concept in calculus. It measures how a function changes as its input changes, essentially providing the rate of change or the slope of the function at any given point.
In this exercise, we started with a derivative:

• Given derivative: \( f'(x) = 2x - 1 \)

The goal was to find the original function before differentiation. The derivative tells us how quickly the function is changing at each point relative to \( x \). When calculating the derivative of \( f(x) = x^2 - x \), we would apply the process known as differentiation to arrive back at \( f'(x) = 2x - 1 \).

Understanding derivatives is crucial for solving many practical problems in physics, engineering, and economics, where determining the rate of change is essential.

Constant of Integration

When we integrate a derivative to find the original function, a crucial step is handling the constant of integration. This constant represents the family of functions that all have the same rate of change, the same derivative, but differ only by a constant.
After integrating, the function appeared as:

• \( f(x) = x^2 - x + C \)

That \( C \) is the constant of integration. It accounts for any vertical shifts in the graph of the function that do not affect the slope represented by the derivative.

To find the specific value of \( C \), more information is required, such as a point through which the function passes. In our exercise, using point \( P(0, 0) \) allowed us to solve for \( C \) by substituting these coordinates,

• \( f(0) = 0^2 - 0 + C = 0 \)

Thus, \( C = 0 \). This step is vital for finding the precise function rather than the family of functions.

Power Rule

The Power Rule is a basic yet powerful tool in calculus for both differentiation and integration. It simplifies the process of finding the derivative or the integral of functions that are polynomial in form.
For differentiation, the Power Rule states:

• If \( f(x) = x^n \), then \( f'(x) = n \cdot x^{n-1} \).

For integration, it is similarly easy and is expressed as:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

In our exercise, we applied the Power Rule for integration to different parts of the expression \( 2x - 1 \):

• Integrating \( 2x \) became \( x^2 \) after applying the rule.
• The constant \(-1\) was integrated as \(-x\).

The Power Rule is straightforward and helps solve many polynomial integration problems, allowing the recreation of the original function from its derivative easily.