Problem 57 - Calculus and its applications

Question

Find f such that: $$ f^{\prime}(x)=\frac{4}{\sqrt{x}}, \quad f(1)=-5 $$

Step-by-Step Solution

Answer

The function is \( f(x) = 8\sqrt{x} - 13 \).

Step 1: Understand the Problem

We are given the derivative of a function, \[ f'(x) = \frac{4}{\sqrt{x}} \]and an initial condition for the function, which is \( f(1) = -5 \). The task is to find the original function \( f(x) \).

Step 2: Integrate the Derivative

To find the original function \( f(x) \), we need to integrate the derivative. This means we need to compute the integral of \( f'(x) \):\[ f(x) = \int \frac{4}{\sqrt{x}} \, dx. \]Recall that \( \sqrt{x} = x^{\frac{1}{2}} \), thus \( \frac{1}{\sqrt{x}} = x^{-\frac{1}{2}} \).

Step 3: Perform the Integration

The integral becomes:\[ \int 4x^{-\frac{1}{2}} \, dx. \]To integrate, use the power rule:\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C. \]Thus, the integration yields:\[ f(x) = 4 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C = 8x^{\frac{1}{2}} + C. \]

Step 4: Use the Initial Condition

We are given that \( f(1) = -5 \). Substitute \( x = 1 \) and \( f(x) = -5 \) into the function:\[ -5 = 8(1)^{\frac{1}{2}} + C. \]This simplifies to:\[ -5 = 8 + C. \]Solving for \( C \), we get:\[ C = -13. \]

Step 5: Write the Final Function

Substitute \( C = -13 \) back into the expression for \( f(x) \):\[ f(x) = 8x^{\frac{1}{2}} - 13. \]This is the original function that satisfies the given derivative and initial condition.

Key Concepts

Integration

In the world of calculus, integration is the process of finding a function given its derivative. Essentially, it is like piecing together a puzzle to figure out the original picture from scattered fragments. When you have the derivative of a function, as noted in the exercise where we have \( f'(x) = \frac{4}{\sqrt{x}} \), integrating this equation allows us to trace back to the original function that was differentiated.

• The process of integration involves reversing differentiation.
• Think of integration as finding the area under a curve.
• Mathematically, it’s denoted by the integral symbol \( \int \).

To integrate, we found \( f(x) = \int \frac{4}{\sqrt{x}} \, dx \). By recognizing \( \frac{4}{\sqrt{x}} \) as \( 4x^{-\frac{1}{2}} \), we simplified the problem into a form that can be solved using the power rule. Once integrated, we obtained a general solution, which includes an arbitrary constant \( C \), since integrating introduces an unknown constant without specific limits.

Initial Conditions

Initial conditions help us find unique solutions to problems involving integration. They act as clues to solve for any constants introduced during integration, providing the exact function needed in real-world applications or given scenarios. Imagine having a treasure map: the initial condition is like that final clue that pinpoints exactly where the 'X' is.In our exercise, the initial condition was given as \( f(1) = -5 \). This means when \( x = 1 \), the function \( f(x) \) equals \(-5\). Without this specific clue:

• We would have many potential functions, all differing only by a constant \( C \).
• The initial condition allows us to solve for \( C \).

By substituting \( x = 1 \) and \( f(x) = -5 \) into the integrated function, we solved for \( C \), ensuring our answer matched the problem context. This pinpointed the exact form of \( f(x) = 8x^{\frac{1}{2}} - 13 \), complying with both the derivative and the initial condition.

Power Rule

The power rule is a handy shortcut in calculus for finding the antiderivative, which is just a fancy word for the function you get after integration. It provides a straightforward way to integrate polynomials and functions that can be expressed in terms of powers of \( x \).The power rule for integration states:- \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \). Here, \( n eq -1 \).Throughout our exercise, the power rule helped simplify and solve the integral \( \int 4x^{-\frac{1}{2}} \, dx \). By increasing the exponent by one, from \(-\frac{1}{2}\) to \(\frac{1}{2}\), and then dividing by this new exponent, we simplified the expression:- The result was \( 8x^{\frac{1}{2}} \), plus the integration constant \( C \).Employing the power rule allows us to transform complex-looking expressions into straightforward ones that are easier to handle, reinforcing its value as a fundamental calculus tool. When used along with initial conditions, it molds the general solution into a specific, functional form.