Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that \(F^{\prime}(x)=5\) and \(F(0)=4\). Find a formula for \(F(x)\). Hint: See Problem 31 .

Step-by-Step Solution

Answer

The formula for \(F(x)\) is \(5x + 4\).

Step 1: Understand the given information

We are provided with the derivative of the function, which is constant: \(F^{\prime}(x) = 5\). Additionally, we know an initial condition or a point: \(F(0) = 4\). Our task is to find the function \(F(x)\).

Step 2: Integrate the derivative

Since \(F^{\prime}(x) = 5\), we integrate the derivative to find the original function \(F(x)\). The integral of 5 with respect to \(x\) is \(5x + C\), where \(C\) is the constant of integration. Therefore, \(F(x) = 5x + C\).

Step 3: Apply the initial condition

We now use the initial condition \(F(0) = 4\) to solve for \(C\). Substitute 0 for \(x\) in the equation \(F(x) = 5x + C\): \[F(0) = 5(0) + C = 4.\] Thus, \(C = 4\).

Step 4: Write the formula for \(F(x)\)

Substitute \(C = 4\) back into the function \(F(x) = 5x + C\). Therefore, the final formula for \(F(x)\) is: \[F(x) = 5x + 4.\]

Key Concepts

Constant of Integration

When integrating a function, finding an antiderivative involves more than just reversing differentiation. After integrating, we often add an arbitrary constant known as the constant of integration, denoted usually by "C". This constant is crucial.

• Integration is a process that "undoes" differentiation. Since the derivative of a constant is zero, when integrating a function, we add this constant to account for any constant term that might have originally been present in the function before it was differentiated.


• It ensures that all possible original functions are represented, as differing only by a constant, they all share the same derivative.


• The constant "C" is determined when additional information, such as an initial condition, is provided (e.g., a specific value of the function at a point).

In the example problem, after integrating the constant derivative of 5, we obtain the general function form: \(F(x) = 5x + C\). The constant "C" is then solved using the initial value given.

Initial Value Problem

An initial value problem (IVP) is a strategy to find a specific solution to a differential equation. This involves both the derivative or derivatives of a function and initial conditions that provide specific information. These conditions help determine the unknown constants.

• The initial value acts as a pivotal point: a known value of the function at a specific point (like \(F(0) = 4\)).


• By substituting the initial values into the integrated expression, we can solve for the constant of integration and thus find the particular solution to the differential equation.

In the problem, given \(F^{\prime}(x) = 5\), the integrated general form of the function is \(F(x) = 5x + C\). We use the initial condition \(F(0) = 4\) to find that \(C = 4\), giving us a precise function \(F(x) = 5x + 4\).

Derivative

The derivative of a function represents its rate of change or the slope of the tangent at any point on the function graph. Derivatives tell us how a function changes as its input changes.

• In simple terms, for every little change in the input \(x\), the output of the derivative tells us how much the output \(F(x)\) changes.


• Derivatives can be constant, as in the problem where \(F^{\prime}(x) = 5\). This means that the function increases at a constant rate of 5 units for every unit increase in \(x\).

Knowing the derivative gives us powerful information to recover the original function by integration. Once the constant of integration is determined using additional conditions, the full behavior of the function is revealed, as shown when deriving \(F(x) = 5x + 4\).