Question

Let f(x) be a function that has a Taylor Series centred at x=6 given by f(x)=∑ n=2[infinity]​ (2n+1)!/3 n n 2 ​ (x−6) ²ⁿ+² converges: What is the smallest value of k for which the derivative f (k) (6)=0?k= What is the value of this derivative? fᵏ (6)=

Answer 1

The question asks for the smallest value of k for which the derivative f^(k)(6) is not equal to 0. Without the exact expression for f(x), we can't determine the value of k or the derivative.

Step-by-step explanation:

The question asks for the smallest value of k for which the derivative f(k)(6) is not equal to 0. To find this value, we need to differentiate the function f(x) with respect to x and evaluate the derivative at x = 6.

To do this, we can start by expressing the Taylor series for f(x) in a more simplified form. Using the formula for the given series, we can rewrite it as:

f(x) = Σ(n=2 to ∞) [(2n+1)! / (3n * n2) * (x-6)2n+2]

Since the question doesn't provide the exact expression for f(x), we can't determine the derivative f(k)(6) or the smallest value of k without further information.