Problem 34 - Calculus Early Transcendentals

Question

a. Find the nth-order Taylor polynomials for the given function centered at the given point a, for \(n=0,1,\) and 2 b. Graph the Taylor polynomials and the function. $$f(x)=\sqrt[3]{x}, a=8$$

Step-by-Step Solution

Answer

Answer: The 2nd-order Taylor polynomial is \(T_2(x) = 2 + \frac{1}{12}(x-8) - \frac{1}{72}(x-8)^2\).

Step 1: Find the first and second derivatives of the given function

For this, we first find the derivative of \(f(x)=\sqrt[3]{x}\). $$f'(x) = \frac{1}{3}\cdot \sqrt[3]{x^{-2}}$$ Now, we find the second derivative of the function: $$f''(x) = \frac{-2}{9}\cdot \sqrt[3]{x^{-3}}.$$

Step 2: Evaluate derivatives at the given point a=8

To find the Taylor polynomials, we need to find the values of the derivatives at the point \(a = 8\). $$f(8) = \sqrt[3]{8} = 2$$ $$f'(8) = \frac{1}{3}\cdot \sqrt[3]{8^{-2}} = \frac{1}{12}$$ $$f''(8) = \frac{-2}{9}\cdot \sqrt[3]{8^{-3}} = \frac{-1}{36}.$$

Step 3: Construct the Taylor polynomials

We use the formula for Taylor polynomials to find the \(n=0,1,2\) order Taylor polynomials. For n=0: $$T_0(x) = f(8) = 2$$ For n=1: $$T_1(x) = 2 + \frac{1}{12}(x-8)$$ For n=2: $$T_2(x) = 2 + \frac{1}{12}(x-8) - \frac{1}{72}(x-8)^2.$$

Step 4: Graph the Taylor polynomials and the function

To graph the function \(f(x)=\sqrt[3]{x}\) and its Taylor polynomials \(T_0(x)\), \(T_1(x)\), and \(T_2(x)\), follow these steps: 1. Plot the original function: graph \(y = \sqrt[3]{x}\) as a curve. 2. Graph the 0th-order Taylor polynomial: plot the horizontal line \(y = 2\). 3. Graph the 1st-order Taylor polynomial: plot the line \(y = 2 + \frac{1}{12}(x-8)\). 4. Graph the 2nd-order Taylor polynomial: plot the curve \(y = 2 + \frac{1}{12}(x-8) - \frac{1}{72}(x-8)^2\). By observing the graphs, we can see how the Taylor polynomials of different orders are approximating the original function around the given point \(a=8\). The higher the order, the better the approximation.

Key Concepts

Derivatives

Derivatives are fundamental in calculus and are crucial in constructing Taylor polynomials. They give us the rate at which a function is changing at any given point. In our exercise, we are working with the function \( f(x) = \sqrt[3]{x} \).
To find its derivatives, we start with the first derivative:

• The first derivative, \( f'(x) \), measures the instantaneous rate of change of the function. For our function, it is \( f'(x) = \frac{1}{3}\cdot \sqrt[3]{x^{-2}} \).
• The second derivative, \( f''(x) \), describes the rate of change of the first derivative, informing us about the concavity of the function. Here, we find \( f''(x) = \frac{-2}{9}\cdot \sqrt[3]{x^{-3}} \).

These derivatives help us understand the behavior of the function at specific points, which is essential when building Taylor polynomials.

Function Approximation

Function approximation involves representing complex functions with simpler polynomials through concepts such as Taylor polynomials.
In our scenario, the goal is to evaluate and construct Taylor polynomials for the function \( f(x) = \sqrt[3]{x} \) at \( a=8 \). Here's how:

• First, we find the function value and derivatives at \( a=8 \). These are given by \( f(8) = 2 \), \( f'(8) = \frac{1}{12} \), and \( f''(8) = \frac{-1}{36} \).
• Using these values, we build the Taylor polynomials:

For \( n=0 \), the function is approximated as a constant: \( T_0(x) = 2 \).
For \( n=1 \), a linear approximation: \( T_1(x) = 2 + \frac{1}{12}(x-8) \).
For \( n=2 \), the quadratic approximation: \( T_2(x) = 2 + \frac{1}{12}(x-8) - \frac{1}{72}(x-8)^2 \).These polynomials approximate \( \sqrt[3]{x} \) near \( x = 8 \) with increasing precision as we increase \( n \).

Graphing

Graphing is an essential tool for visual understanding of how Taylor polynomials approximate functions. In this case, it allows us to compare \( \sqrt[3]{x} \) and our Taylor polynomials around \( x = 8 \).To begin:

• Graph the original function, \( f(x) = \sqrt[3]{x} \), to see the curve's general behavior.
• Plot \( T_0(x) = 2 \): a horizontal line depicting the most basic approximation.
• Next, graph \( T_1(x) = 2 + \frac{1}{12}(x-8) \), which shows a linear angle of approach near \( x = 8 \).
• Finally, draw \( T_2(x) = 2 + \frac{1}{12}(x-8) - \frac{1}{72}(x-8)^2 \), offering a more nuanced fit with curvature.

By observing these graphs, it's clear how the Taylor polynomials tend to closely follow the function as the polynomial order increases. It's a visual testament to the power of derivatives and Taylor polynomials in approximating functions.