Problem 21 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Based on the step-by-step solution provided, write the nth-order Taylor polynomial for the function f(x) = (1+x)^(-3) centered at x = 0 for n = 0, 1, and 2.

Step 1: Identify the function and compute derivatives

Given function is: $$ f(x)=(1+x)^{-3} $$ Now, compute the first few derivatives of the function: $$ f'(x)=-3(1+x)^{-4}(1)=-3(1+x)^{-4} $$ $$ f''(x)=12(1+x)^{-5}(1)=12(1+x)^{-5} $$

Step 2: Evaluate the derivatives at x = 0

Now evaluate the derivatives at x = 0: $$ f(0)=(1+0)^{-3}=1 $$ $$ f'(0)=-3(1+0)^{-4}=-3 $$ $$ f''(0)=12(1+0)^{-5}=12 $$

Step 3: Find Taylor polynomials for n=0,1, and 2

Use the Taylor polynomial formula stated in the Analysis section for each n: For n=0: $$ T_{0}(x)=\frac{f(0)}{0!}(x)^{0}=1 $$ For n=1: $$ T_{1}(x)=\frac{f(0)}{0!}(x)^{0}+\frac{f'(0)}{1!}(x)^{1}=1-3x $$ For n=2: $$ T_{2}(x)=\frac{f(0)}{0!}(x)^{0}+\frac{f'(0)}{1!}(x)^{1}+\frac{f''(0)}{2!}(x)^{2}=1-3x+\frac{12}{2}x^{2}=1-3x+6x^{2} $$

Step 4: Graph the function and the Taylor polynomials

Now graph the function and the Taylor polynomials \(T_{0}(x), T_{1}(x),\) and \(T_{2}(x)\) on the same plot. The student can use graphing software or calculator to plot them. In summary, the nth-order Taylor polynomials for given function are: - For n=0: \(T_{0}(x)=1\) - For n=1: \(T_{1}(x)=1-3x\) - For n=2: \(T_{2}(x)=1-3x+6x^{2}\) Now graph the Taylor polynomials and the function to visualize the approximations.