Problem 52 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find \(f(x)\) by solving the initial value problem. $$f^{\prime}(x)=3 x^{2}-6 x ; f(2)=4$$

Step-by-Step Solution

Answer

The function \(f(x)\) that satisfies the given initial value problem is \(f(x) = x^3 - 3x^2 + 8\).

Step 1: Integrate \(f'(x)\) to find an expression for \(f(x)\)

To find the function \(f(x)\), we need to integrate its derivative \(f'(x)\). So, let's integrate \(f'(x) = 3x^2 - 6x\): \[ \int{(3x^2 - 6x) dx} = \int{3x^2 dx} - \int{6x dx} \] Now integrate each term separately: \[ \int{3x^2 dx} = x^3 + C_1 \\ \int{6x dx} = 3x^2 + C_2 \] Adding these two integrals, we get: \[ f(x) = x^3 - 3x^2 + C \] where \(C = C_1 - C_2\).

Step 2: Use the initial condition to find the value of the constant of integration

Now, we will use the initial condition \(f(2) = 4\) to find the value of \(C\). Plug \(x = 2\) into the expression for \(f(x)\): \[ f(2) = (2)^3 - 3(2)^2 + C \] Using the initial condition \(f(2) = 4\), we can write \[ 4 = 8 - 12 + C \] Solving for \(C\), we get \(C = 8\).

Step 3: Write the final expression for the function \(f(x)\)

Now that we have found the constant of integration \(C\), we can write the final expression for the function \(f(x)\): \[ f(x) = x^3 - 3x^2 + 8 \] So the function \(f(x)\) that satisfies the given initial value problem is \(f(x) = x^3 - 3x^2 + 8\).

Key Concepts

Initial Value Problem

In calculus, solving an initial value problem (IVP) is a process where we're looking for a function that satisfies both a differential equation and an initial condition. The differential equation gives us the rate of change of a function, while the initial condition gives us an exact point through which this function passes.

For example, in our given exercise, we have the derivative of a function, \(f'(x) = 3x^2 - 6x\). To find the function \(f(x)\), we integrate \(f'(x)\), using the initial condition \(f(2) = 4\) to determine the unique solution. Problems like this require understanding both integration and the application of specific conditions to pin down the exact form of the function.

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. They describe how a function changes and are particularly useful in modeling real-world systems. This mathematical tool helps us express relationships involving rates of change and physical quantities.

In our original problem, the differential equation is \(f'(x) = 3x^2 - 6x\). This tells us that the function's rate of change is modeled by \(3x^2 - 6x\), which represents a varying slope at each point along the curve of the original function. Solving this involves finding a function \(f(x)\) that will have a derivative equal to this equation.

Integration

Integration is the process of finding the original function from its derivative. It's essentially the reverse of differentiation. While differentiation gives you the slope or rate of change at any given point, integration allows you to "accumulate" these changes to discover the original function itself.

In the exercise, integrating the derivative function \(f'(x) = 3x^2 - 6x\) was performed as follows:

• The term \(3x^2\) was integrated to become \(x^3 + C_1\).
• The term \(-6x\) was integrated to become \(-3x^2 + C_2\).

The result was the expression \(f(x) = x^3 - 3x^2 + C\), where \(C\) is the combined constant from the integration of each term. This expression represents the general solution of the differential equation.

Constant of Integration

When integrating, a constant of integration must be added because differentiation loses constant information. This means when you find the integral of a derivative, you need to add '+ C' to represent any constant value that may have been present in the original function but lost during differentiation.

For the problem at hand, after integrating \(f'(x) = 3x^2 - 6x\), we added a constant \(C\) to write \(f(x) = x^3 - 3x^2 + C\). To find this constant, we used the initial condition \(f(2) = 4\). By substituting \(x = 2\) into the integrated function and setting it equal to 4, we solved for \(C\), finding \(C = 8\).

The constant ensures that the solution to the differential equation is unique and satisfies the given initial value.