Problem 35 - Calculus Single Variable

Question

In each of Exercises \(35-40\), an expression \(f(x)\) is given. Find all functions \(F\) such that \(F^{\prime}(x)=f(x)\). $$ 5 $$

Step-by-Step Solution

Answer

The function \( F(x) = 5x + C \) satisfies \( F'(x) = 5 \).

Step 1: Understanding the Problem

We need to find a function \( F(x) \) whose derivative is \( f(x) = 5 \). This involves finding the antiderivative or integral of 5 with respect to \( x \).

Step 2: Integrate the Constant Function

The derivative of a constant \( k \) is zero, and the derivative of \( x \) with a constant multiplied is the constant itself. Recognizing this, the function \( f(x) = 5 \) can be integrated as follows: \[ F(x) = \int 5 \, dx = 5x + C \] where \( C \) represents the constant of integration.

Step 3: Verification

Verify that the derivative of \( F(x) = 5x + C \) is indeed \( 5 \). \[ \frac{d}{dx}(5x + C) = 5 \cdot \frac{d}{dx}(x) + \frac{d}{dx}(C) = 5 \cdot 1 + 0 = 5 \] This confirms our solution is correct.

Key Concepts

Antiderivative

The concept of an antiderivative is central to the exercise at hand. When we talk about an antiderivative, we're referring to the reverse process of differentiation. In simple terms, if the derivative of a function gives us a specific expression, finding the antiderivative means finding a function that, when differentiated, returns the original expression. For example, if we know that the derivative of some function is 5, we want to determine the original function from which this derivative was obtained.

Finding an antiderivative can also be termed as integration. In this exercise, we want to integrate the constant function 5. The result is expressed as an integral:

• The integral of a constant 'c' with respect to a variable, say 'x', is simply 'c' times that variable. Hence, \[ \int 5 \, dx = 5x + C \]
• The "+ C" signifies that there are infinite possible functions, each differing by a constant.

This means that all functions of the form \( F(x) = 5x + C \) will have a derivative of 5, regardless of the value of \( C \).

Constant of Integration

When dealing with integration, one term often appears: the constant of integration, denoted by \( C \). This concept is important because when taking antiderivatives, we need to acknowledge that there are potentially infinite solutions. During differentiation, the derivative of any constant is zero, meaning that any constant present in an initial function doesn't impact its derivative.

Thus, when reversing the process from a derivative to an antiderivative, it is vital to add this constant \( C \). Consider this:

• If \( F(x) = 5x \), then \( F'(x) = 5 \).
• Equally, if \( F(x) = 5x + 7 \), then \( F'(x) \) is still 5.
• The 7 here is arbitrary, showing that our antiderivative can differ by any constant.

Therefore, whenever an indefinite integral is evaluated, including the constant of integration ensures the complete set of solutions.

Derivative

The derivative is a fundamental concept in calculus and relates directly to how a function changes as its input changes. In notation, if you have a function \( F(x) \), its derivative is denoted \( F'(x) \). Taking the derivative involves calculating the rate of change or the slope of the function \( F(x) \) at any given point.

In our example exercise, \( F'(x) = f(x) \) is given as 5, which tells us that the rate of change of the function \( F(x) \) is constant across all \( x \)-values:

• This implies that the slope of the tangent line to \( F(x) \) is horizontally constant at all points.
• A constant derivative typically indicates a linear function, which our antiderivative \( 5x + C \) confirms.

Therefore, reversing this derivative process by taking an antiderivative or integral of \( f(x) \), provides us with the original function within which a constant of integration must be included to capture every possible original state of the function.