Problem 16 - University Calculus: Early Transcendentals

Question

Use the linear approximation \((1+x)^{k} \approx 1+k x\) to find an approximation for the function \(f(x)\) for values of \(x\) near zero. a. \(f(x)=(1-x)^{6}\) b. \(f(x)=\frac{2}{1-x}\) c. \(f(x)=\frac{1}{\sqrt{1+x}}\) d. \(f(x)=\sqrt{2+x^{2}}\) e. \(f(x)=(4+3 x)^{1 / 3}\) \(f(x)=\sqrt[3]{\left(1-\frac{x}{2+x}\right)^{2}}\)

Step-by-Step Solution

Answer

a. \(f(x) \approx 1 - 6x\); b. \(f(x) \approx 2 + 2x\); c. \(f(x) \approx 1 - \frac{1}{2}x\); d. \(f(x) \approx 1 + \frac{x^2}{4}\); e. \(f(x) \approx 4^{1/3}(1 + \frac{x}{4})\); f. \(f(x) \approx 1 - \frac{x}{2+x}\).

Step 1: Identify the function form

We need to express each function in the form \((1+x)^k\). This allows us to use the given linear approximation formula \((1+x)^{k} \approx 1+kx\).

Step 2a: Approximate \(f(x)=(1-x)^6\)

Rewrite \(f(x)\) as \((1+(-x))^6\). Using the linear approximation for \(x\) near zero, we have:\(f(x) \approx 1 + 6(-x) = 1 - 6x\).

Step 2b: Approximate \(f(x)=\frac{2}{1-x}\)

Rewrite \(f(x)\) as \(2(1-x)^{-1}\). Using the approximation \((1+x)^{-1} \approx 1-x\), we have:\(f(x) \approx 2(1+(-x)) = 2(1+x) = 2 + 2x\).

Step 2c: Approximate \(f(x)=\frac{1}{\sqrt{1+x}}\)

Rewrite \(f(x)\) as \((1+x)^{-1/2}\). Applying the approximation, we get:\(f(x) \approx 1 - \frac{1}{2}x\).

Step 2d: Approximate \(f(x)=\sqrt{2+x^{2}}\)

Set the function as \((1+\frac{x^2}{2})^{1/2}\). Applying the approximation:\(f(x) \approx 1 + \frac{1}{2}\cdot\frac{x^2}{2} = 1 + \frac{x^2}{4}\).

Step 2e: Approximate \(f(x)=(4+3x)^{1/3}\)

Express the function as \((4)^{1/3}(1+\frac{3x}{4})^{1/3}\). Approximate as:\(f(x) \approx 4^{1/3}(1 + \frac{3x}{12}) = 4^{1/3}(1 + \frac{x}{4})\).

Step 2f: Approximate \(f(x)=\sqrt[3]{\left(1-\frac{x}{2+x}\right)^{2}}\)

Rewrite the function as \((1+(-\frac{x}{2+x}))^2)^{1/3}\). Approximate this as:\(f(x) \approx (1 + 2(-\frac{x}{2+x}))^{1/3} \approx 1 - \frac{x}{2+x}\).

Key Concepts

Taylor Series

The Taylor Series is a powerful tool in calculus used for approximating functions around a certain point. This series allows us to express complex functions as infinite sums of their derivatives evaluated at a specific point. For instance, if we want to approximate a function \(f(x)\) around \(x = a\), the Taylor Series is given by:
\[f(x) \approx f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots\]When \(x\) is near \(a\), the terms involving higher powers of \(x-a\) contribute less to the sum, allowing us to truncate the series for a quick approximation.
In the context of our problem, we used a form of the first order Taylor approximation, which can be seen in linear approximation methods like \((1+x)^k \approx 1+kx\). This is because we're looking at the approximation for values of \(x\) near zero, making higher power terms negligible for small \(x\).

• It's quick: By stopping at just the first derivative, we make the approximation fast.
• It's simple: Linear approximations simplify calculus concepts to basic algebra, suitable for small changes.

Approximation Methods

Approximation methods are techniques used to estimate the value of mathematical functions that might be difficult or impossible to compute exactly. These methods help bridge the gap between theoretical mathematics and practical computations.
Linear approximation, one of the simplest forms, uses the tangent line at a point to approximate the function near that point. The formula \((1+x)^k \approx 1+kx\) is a perfect example. It's derived from the idea of taking the linear component of the Taylor series and ignoring higher degree terms.

• Useful for small values of \(x\): They give a reasonable estimation when \(x\) is close to zero.
• Widely applied: These are frequently used in engineering and other sciences for quick calculations.

These methods provide a way to quickly gauge the behavior of functions, especially when high precision is not critical.

Calculus Concepts

Calculus is the study of change, and it provides tools to understand and describe changes in functions. Basic calculus concepts such as derivatives and integrals form the building blocks for understanding how functions behave.
In this exercise, we specifically utilize differentiation, a core concept in calculus, to develop our linear approximations. The derivative of a function gives the rate at which the function's value changes at any given point, which is why it's central to constructing linear approximations.

• Derivatives: Understanding how to compute and apply them is essential for approximations.
• Tangents: The tangent line at a point provides the best linear approximation to the function at that point.

These concepts work together to allow us to model and predict behavior in a framework that's both mathematically rich and practically insightful.