Problem 13 - Calculus with Concepts in Calculus

Question

Determine the function \(\mathrm{f}\) satisfying the given conditions. $$ f^{\prime}(x)=-2, f(0)=0 $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to find a function \( f(x) \) whose derivative is \( f'(x) = -2 \) and that passes through the point \((0,0)\). This means we should find \( f(x) \) based on the given derivative and initial condition.

Step 2: Integrate the Derivative

To find \( f(x) \), integrate the derivative \( f'(x) = -2 \). The integral of \(-2\) with respect to \( x \) is \(-2x + C\), where \( C \) is the constant of integration.\[ f(x) = -2x + C \]

Step 3: Apply the Initial Condition

Use the initial condition \( f(0) = 0 \) to find \( C \). Substitute \( x = 0 \) and \( f(x) = 0 \) into the equation \( f(x) = -2x + C \).\[ 0 = -2(0) + C \]This simplifies to \( C = 0 \).

Step 4: Write the Final Function

Substitute \( C = 0 \) back into the integrated function to find the final expression for \( f(x) \).\[ f(x) = -2x + 0 \]Thus, the function is \( f(x) = -2x \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that allows us to find the rate at which a function is changing at any given point. In simpler terms, it tells us how a function behaves or evolves. The derivative of a function, often denoted as \( f'(x) \) or \( \frac{df}{dx} \), gives the slope of the tangent line to the function at any point. For a linear function such as \( f(x) = -2x \), the derivative is constant: \( f'(x) = -2 \). This means the rate of change, or slope, is always -2, indicating a uniformly decreasing function.

When working with more complex functions, such as polynomials, exponentials, or trigonometric functions, differentiation allows us to calculate instantaneous rates of change. This can be particularly useful in physics for finding velocity from position functions, or in economics for finding the marginal cost from a cost function.

Understanding how to differentiate various types of functions helps to solve various real-world problems by analyzing how changing one variable influences another. Differentiation is a powerful tool that opens up many pathways to solving problems in science, engineering, and beyond.

Integration

Integration is the process of finding the original function when its derivative is known; it is essentially the reverse of differentiation. When we integrate a function, we're often looking for the area under the curve the function creates, or we're reconstructing a function from its derivative. In the current exercise, this meant finding the function \( f(x) \) from its derivative \( f'(x) = -2 \).

To integrate \( f'(x) = -2 \), we find the antiderivative, which is \(-2x + C\). Here, \( C \) is the constant of integration, representing any fixed output shift in the function. This constant is vital because the derivative of any constant is zero, so all functions of the form \(-2x + C\) will have the same derivative \( -2 \).

Integration is especially useful in various fields like physics, where it can calculate quantities like displacement when given a velocity function, or total accumulation from rate of change. It is a central tool in calculus for solving problems where the total effect over time or space needs to be determined.

Initial Value Problem

An initial value problem (IVP) in calculus involves finding a function that satisfies a differential equation and also meets a specific starting condition. The exercise we examined requires solving an IVP by finding the function \( f(x) \) with the derivative \( f'(x) = -2 \) and passing through the point \((0,0)\).

This involves two main steps: first, integrate the derivative to find the general form of the function, which we did to obtain \( f(x) = -2x + C \). Then, apply the initial condition to determine the constant \( C \). By substituting the coordinates of the known point \((0,0)\) into \( f(x) = -2x + C \), we found that \( C = 0 \), giving the specific solution \( f(x) = -2x \).

IVPs are crucial when you require an exact solution, not only for science and engineering calculations where the function must fulfill certain criteria at initial conditions, but also in economics where initial conditions can significantly influence future predictions. Understanding how to solve these problems help students piece together comprehensive solutions from basic principles.