Question

Question: In Problems 29–34, determine the Taylor series about the point X₀for the given functions and values of X₀.

33. f (X)= x³+3x-4, x₀= 1

Short answer

The required expression is 6(x-1)+3(x-1)²+(x-1)³.

Step-by-step solution

Taylor series

For a function f(x) the Taylor series expansion about a point x₀is given by,$f(x-x_0)=f\left(x_0\right)+f\text{'}\left(x_0\right).(x-x_0)+f\text{'}\text{'}\left(x_0\right).\frac{{(x-x_0)}^2}{2!}+f\text{'}\text{'}\text{'}\left(x_0\right).\frac{{(x-x_0)}^3}{3!}+$.....

Derivatives of function at x0.

We have to calculate the Taylor series expansion for, f (x) = x³+3x-4 at x₀=1 .

Calculating the derivatives of function at x₀

f(x) = x³+3x-4 then ^{f (X₀)=0}

f'(x) = 3x²+3 then f'(x₀)=6

f''(x) = 6x then f''(x₀)=6

f'''(x) =6 thenf'''(x₀)=6

f''''(x)=0 thenf''''(x₀)=60

Substitute the derivatives in Taylor series

Substituting the above derivatives in Taylor series expansion for the function at ,

x₀=1 .then,

$$\begin{gathered} x^3+3x-4=0+6.(x-1)+6.\frac{{\left(x-1\right)}^2}{2!}+6.\frac{{\left(x-1\right)}^3}{3!}+0+0+.... \\ =6(x-1)+3{(x-1)}^2+{(x-1)}^3 \end{gathered}$$

Hence, the required expression is 6(x-1)+3(x-1)²+(x-1)³