Problem 30 - Calculus for Biology and Medicine

Question

Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=\left(1+\frac{1}{x}\right)^{1 / 4}\) at \(a=1\)

Step-by-Step Solution

Answer

The linear approximation is \( L(x) = 2^{1/4} - \frac{1}{4} \cdot 2^{-3/4} (x - 1) \).

Step 1: Define the Function and the Point

We are given the function \( f(x) = \left(1 + \frac{1}{x}\right)^{1/4} \) and need to find its linear approximation at \( a = 1 \). This means we will calculate \( f(1) \) and \( f'(1) \).

Step 2: Evaluate the Function at the Point

Substitute \( a = 1 \) into the function: \[ f(1) = \left(1 + \frac{1}{1}\right)^{1/4} = 2^{1/4}. \]

Step 3: Find the Derivative of the Function

Use the chain rule to differentiate \( f(x) = \left(1 + \frac{1}{x}\right)^{1/4} \). Let \( u(x) = 1 + \frac{1}{x} \), so \( f(x) = u(x)^{1/4} \). The derivative is: \[ f'(x) = \frac{1}{4}u(x)^{-3/4} \cdot u'(x). \]

Step 4: Differentiate the Inner Function

Find \( u'(x) \): \[ u(x) = 1 + \frac{1}{x}, \, \text{thus} \, u'(x) = -\frac{1}{x^2}. \]

Step 5: Substitute Back into the Derivative

Substitute \( u(x) \) and \( u'(x) \) into the derivative: \[ f'(x) = \frac{1}{4}\left(1 + \frac{1}{x}\right)^{-3/4} \left(-\frac{1}{x^2}\right) = -\frac{1}{4x^2}\left(1 + \frac{1}{x}\right)^{-3/4}. \]

Step 6: Evaluate the Derivative at the Point

Evaluate \( f'(1) \) by substituting \( x = 1 \): \[ f'(1) = -\frac{1}{4(1^2)}(2)^{-3/4} = -\frac{1}{4} \cdot 2^{-3/4}. \]

Step 7: Linear Approximation Formula

The linear approximation formula is given by: \[ L(x) = f(a) + f'(a)(x - a). \] Substitute the values for \( f(1) \) and \( f'(1) \): \[ L(x) = 2^{1/4} - \frac{1}{4} \cdot 2^{-3/4} (x - 1). \]

Key Concepts

Differentiation

Differentiation is at the heart of calculus, offering a method to determine the rate at which a function is changing at any given point. This is surprisingly useful, from optimizing systems to calculating slopes.
For a function like \( f(x) \), the derivative \( f'(x) \) represents the instant rate of change of \( f \) concerning \( x \).

• The basic tools of differentiation include power, product, quotient, and chain rules.
• They provide a way to systematically break down complex functions.

In this exercise, we started with the function \( f(x) = \left(1 + \frac{1}{x}\right)^{1/4} \). By differentiating it, we can understand how the output of the function changes as the input variable \( x \) varies.

Chain Rule

The chain rule is an essential technique for differentiation, especially when dealing with composite functions.
This rule allows us to differentiate a function that is made up of other simpler functions.

• For example, if we have a function \( g(x) = (h(x))^n \), we can apply the chain rule to find its derivative.

In the context of this problem, you saw the chain rule in action. The function \( f(x) = \left(1 + \frac{1}{x}\right)^{1/4} \) can be considered as \( u(x)^{1/4} \), where \( u(x) = 1 + \frac{1}{x} \).
To find \( f'(x) \), we differentiated \( u(x)^{1/4} \) by multiplying by the derivative of the inner function \( u(x) \).
This kind of differentiation reveals how each part of a function contributes to its overall rate of change.

Derivative Evaluation

Once a derivative is found, evaluating it at specific points provides insights into a function's behavior at those points.
In this exercise, after deriving \( f'(x) \), we evaluated it at \( x = 1 \). This involved plugging \( x = 1 \) into the derived expression to obtain \( f'(1) \).

• This step is crucial for linear approximation since \( f'(a) \) contributes directly to the slope of the tangent line at \( x = a \).
• In our case, it meant substituting into the expression \(-\frac{1}{4x^2}\left(1 + \frac{1}{x}\right)^{-3/4}\) and simplifying.

Understanding how to evaluate derivatives helps in many areas, including plotting functions and solving real-world problems.

Taylor Series

A Taylor series provides a way to approximate complex functions using polynomials. Normally centered at a specific point, these approximations become more accurate as more terms are included.
The linear approximation is essentially the first term in a Taylor series expansion at a given point \( a \).

• It is expressed as \( f(a) + f'(a)(x-a) \), which uses the function value and the slope to form a tangent line approximation.
• Linear approximations are incredibly useful when the behavior of a function near a point needs to be understood without more complex calculations.

In our scenario, we used this method to approximate the function \( f(x) = \left(1 + \frac{1}{x}\right)^{1/4} \) near \( x = 1 \), allowing us to predict its output with only the function and its derivative's values at that single point.