Question

1 point
What is the Taylor series expansion of f(x) = x2-x+1 about the point x
-1?​

Answer 1

Answer:

3-3(x+1)+(2/2!)(x+1)²

Step-by-step explanation:

taylor expansion at a is

f(x)=f(a)+f'(a)(x-a)+(2/2!)f"(a)(x-a)²+......

f'(x)=2x-1=> f'(a)=-3

f"(x)=2

f"'(x)=0

so

f(x) at x=-1

is

3-3(x+1)+(2/2!)(x+1)²

Answer 2

Answer:

The Taylor series expansion of f(x) = x²-x+1 about the point x = -1 is f(x) = x² - x + 1.

Step-by-step explanation:

Given:

f(x) = x²-x+1

To find:

Taylor expansion of the given function at x = -1

Solution:

The given function is f(x) = x²-x+1

Substituting x = -1 in the function, we get

f(-1) = (-1)²-(-1)+1

f(-1) = 1 + 1 + 1

f(-1) = 3

Taking the derivation of f(x).

f'(x) = 2x - 1

Now, substituting x = -1 in f'(x)

f'(-1) = 2(-1) - 1

f'(-1) = -2 - 1

f'(-1) = -3

Taking the derivative of f'(x).

f''(x) = 2

Therefore, f''(-1) = 2

The Taylor series expansion of f(x) about x=a is,

f(x) =  f(a)+(x-a) f'(a)+(x-a)² f”(a)/2!+⋯

Here, a = -1. Hence, substituting the calculated values, we get

f(x) = f(-1) + (x+1)f'(-1) + (x+1)²f''(-1)/2!

f(x) = 3 + (x+1)(-3) + (x+1)²(2)/2

f(x) = 3 - 3(x+1) + (x+1)²

f(x) = 3 - 3x - 3 + x² + 2x + 1

f(x) = x² - x + 1

Therefore, the Taylor expansion of the given function is f(x) = x² - x + 1.

#SPJ3