Problem 24 - Essential Calculus Early Transcendentals

Question

Find \(f\) $$f^{\prime}(x)=5 x^{4}-3 x^{2}+4, \quad f(-1)=2$$

Step-by-Step Solution

Answer

The function is \(f(x) = x^5 - x^3 + 4x + 6\).

Step 1: Understand the Exercise

We are given the derivative of a function, \(f'(x) = 5x^4 - 3x^2 + 4\), and a condition \(f(-1) = 2\). We need to find the original function \(f(x)\).

Step 2: Integrate to Find the General Form of f(x)

To find \(f(x)\), integrate \(f'(x)\). So, compute \[\int (5x^4 - 3x^2 + 4) \, dx\]. This results in \(f(x) = x^5 - x^3 + 4x + C\), where \(C\) is the integration constant.

Step 3: Apply the Initial Condition

We know \(f(-1) = 2\), which gives us \(f(-1) = (-1)^5 - (-1)^3 + 4(-1) + C = 2\). Simplifying, we get \(-1 + 1 - 4 + C = 2\), so \(C = 6\).

Step 4: Write the Particular Solution

Substitute \(C = 6\) back into \(f(x)\). Therefore, \(f(x) = x^5 - x^3 + 4x + 6\).

Key Concepts

Integration

Integration is a fundamental concept in calculus, often used to determine the original function given its derivative. To find an antiderivative or integral, you essentially reverse the differentiation process.

In the given exercise, the derivative function is provided and you need to find the antiderivative to obtain the original function. By integrating the derivative function, you recover the original function with an additional constant, often represented as "C." This constant arises because derivatives of constant numbers are zero, meaning during differentiation they disappear. Thus, during integration, you reintroduce one as the constant of integration.

For example, if you have a derivative like:

• \( f'(x) = 5x^4 - 3x^2 + 4 \)

You integrate it as:

• \[ f(x) = \int (5x^4 - 3x^2 + 4) \, dx \]

This results in:

• \( f(x) = x^5 - x^3 + 4x + C \)

This reinforces how integration helps in finding the overall function by determining the cumulative area under the curve defined by the derivative.

Initial Conditions

Initial conditions play a crucial role in problems involving antiderivatives. They allow you to solve for the constant of integration, "C," to find a specific solution.

In the problem presented, the initial condition is given as \(f(-1) = 2\). This allows you to substitute x with -1 and solve for "C" in the integrated function. By applying this, the generic function becomes a particular solution exactly fitting the initial constraints.

After integrating, we found that:

• \( f(x) = x^5 - x^3 + 4x + C \)

Using the initial condition, substitute x and f(x) as follows:

• \( (-1)^5 - (-1)^3 + 4(-1) + C = 2 \)

Simplifying the equation yields:

• \( -1 + 1 - 4 + C = 2 \)

Thus:

• \( C = 6 \)

This calculation illustrates how initial conditions are used to pinpoint the exact form of the function needed.

Power Rule in Calculus

The power rule is a key tool for both differentiation and integration in calculus. It provides a simple way to handle functions of the form \(x^n\). For integration, the power rule states:

• \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]

where \(n\) is not equal to -1. In our problem, we used the power rule to integrate each term in the derivative function.

Consider the problem's function:

• \(5x^4 - 3x^2 + 4\)

Using the power rule, each term is integrated separately:

• For \(5x^4\), applying the power rule gives \( \frac{5}{5}x^5 = x^5 \)
• For \(-3x^2\), applying the power rule gives \( -1x^3 = -x^3 \)
• For \(4\), applying the power rule gives \( 4x \) (since \(\int 1 \, dx = x\) and you treat "4" as "4(1)")

These integrations provide the components needed to build the formula for the antiderivative. This rule is essential for efficiently solving these types of calculus problems and forming a strong foundational understanding of integrations.