What is the quadratic Taylor polynomial about 2 for the function f(x)=square root x-1?
To find the quadratic Taylor polynomial about 2 for the function f(x) = sqrt(x-1), we first need to find the polynomial's coefficients using the Taylor series expansion.
The Taylor series expansion for a function f(x) centered around a point "a" is given by:
f(x) = f(a) + f'(a)(x - a) + (f''(a)/2!)(x - a)^2 + ...
To find the quadratic Taylor polynomial, we will need to calculate the first and second derivatives of f(x).
First derivative:
f'(x) = (1/2√(x - 1))
Second derivative:
f''(x) = -1/(4(x - 1)^(3/2))
Now, let's evaluate these derivatives at x = 2 to find the coefficients:
f'(2) = (1/2√(2 - 1)) = (1/2)
f''(2) = -1/(4(2 - 1)^(3/2)) = -1/4
The quadratic Taylor polynomial centered around x = 2 is given by:
P(x) = f(2) + f'(2)(x - 2) + (f''(2)/2!)(x - 2)^2
To find f(2), substitute x = 2 into the original function:
f(2) = sqrt(2 - 1) = sqrt(1) = 1
Plug in the values for f(2), f'(2), and f''(2) into the polynomial:
P(x) = 1 + (1/2)(x - 2) + (-1/4)(x - 2)^2
Simplifying the expression, we get:
P(x) = 1/4(x - 2)^2 + 1/2(x - 2) + 1
Therefore, the quadratic Taylor polynomial about 2 for the function f(x) = sqrt(x-1) is P(x) = 1/4(x - 2)^2 + 1/2(x - 2) + 1.