Problem 24 - Calculus Single Variable

Question

Use the method of increments to estimate the value of \(f(x)\) at the given value of \(x\) using the known value \(f(c)\) $$ f(x)=\sqrt{x} /(1+\sqrt{x}), c=9, x=8.6 $$

Step-by-Step Solution

Answer

Estimated value of \(f(8.6)\) is \(\frac{59}{80}\).

Step 1: Identify Known Values

We know that the function is given by \(f(x) = \frac{\sqrt{x}}{1+\sqrt{x}}\). The point \(c = 9\) is where we know the exact function value, \(f(c)\), and we want to estimate \(f(8.6)\) where \(x = 8.6\).

Step 2: Calculate \(f(c)\)

Substitute \(c = 9\) into the function: \[ f(9) = \frac{\sqrt{9}}{1+\sqrt{9}} = \frac{3}{1 + 3} = \frac{3}{4} \].

Step 3: Find the Increment \( Delta x\)

Calculate the increment in \(x\) as \(\Delta x = x - c = 8.6 - 9 = -0.4\).

Step 4: Compute \(f'(x)\) (the Derivative)

First, use the quotient rule to differentiate: \[ f'(x) = \frac{(1+\sqrt{x})(\frac{1}{2\sqrt{x}}) - \sqrt{x}(\frac{1}{2\sqrt{x}})}{(1+\sqrt{x})^2} \]. Simplifying the expression yields \[ f'(x) = \frac{1}{2(1+\sqrt{x})^2} \].

Step 5: Calculate \(f'(c)\)

Substitute \(c = 9\) into the expression for the derivative to find \(f'(c)\): \[ f'(9) = \frac{1}{2(1 + 3)^2} = \frac{1}{32} \].

Step 6: Compute \(\Delta f\)

Calculate the change in function value, \(\Delta f\), using \(\Delta f \approx f'(c) \cdot \Delta x\): \(\Delta f \approx \frac{1}{32} \cdot (-0.4) = -\frac{1}{80}\).

Step 7: Estimate \(f(x)\) at \(x = 8.6\)

Use the approximation \(f(x) = f(c) + \Delta f\): \[ f(8.6) \approx \frac{3}{4} - \frac{1}{80} = \frac{60}{80} - \frac{1}{80} = \frac{59}{80} \].

Key Concepts

Understanding the Quotient Rule

When we need to differentiate a function that is a ratio of two other functions, we use the quotient rule. It helps us find the derivative of expressions where one function is divided by another. For example, if we have a function defined as \(f(x) = \frac{u(x)}{v(x)}\), then the derivative \(f'(x)\) can be calculated using the formula:\[ f'(x) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2} \]This rule is particularly handy in physics and engineering, where such ratios are frequent. Breaking it down:

• Differentiate the numerator \(u(x)\)
• Differentiate the denominator \(v(x)\)
• Apply them in the formula to get \(f'(x)\)

This method ensures we account for the changes in both the top and bottom functions as they alter.

The Power of Derivatives

In calculus, a derivative represents how a function changes as its input changes. Think of it as the function's slope at any point. This concept is fundamental, as it provides insights into trends and behaviors of different functions, such as concavity and rate of change.For example, if you have a function that describes a car's position over time, its derivative tells you the car's speed at any instant. Similarly, in our exercise, finding \(f'(x)\) at a specific point helps us understand how our function \(f(x)\) behaves around \(x = 9\).

Calculating the Derivative

In the provided function, we applied the quotient rule to find the derivative, \(f'(x)\), which tells us how fast the function \(f(x) = \frac{\sqrt{x}}{1+\sqrt{x}}\) is changing at any given point.

What is Function Increment?

A function increment refers to the change in the function's value as its input changes. In mathematical terms, this is expressed as \(\Delta f = f(x + \Delta x) - f(x)\). When the change in input is small, \(\Delta f\) can approximate using its derivative via \(\Delta f \approx f'(x) \cdot \Delta x\).

Why Use Function Increment?

It helps us estimate values of complex functions at points close to where we already have known values. This is because derivatives provide a linear approximation, which is often sufficient for small changes. Hence, in our exercise, we are estimating \(f(8.6)\) from \(f(9)\) using this principle.

Estimating a Function's Value

To estimate a function's value near a known point, we use one of the fundamental ideas in calculus—linear approximation. This involves calculating the derivative at a known point and using it to approximate the function's value close by.In the exercise, we started with a known value \(f(9)\). Using the derivative \(f'(x)\) calculated at \(x = 9\), and the small increment \(\Delta x = -0.4\), we estimated \(f(8.6)\).

Practical Steps

• Find the function value \(f(c)\) you know.
• Calculate the derivative \(f'(c)\) at that point.
• Determine the increment \(\Delta x\) from the known point.
• Estimate the new function value: \(f(x) = f(c) + f'(c) \cdot \Delta x\).

This process simplifies estimating values without computing the function from scratch, especially useful for difficult or complex functions.