Question

How do you calculate f''(1) and f'''(1) given `f(x)= 3- 6(x-1) + 4/(2!) (x-1)^2 - 5/(3!) (x-1)^3 + 1/(4!) (x-1)^4`?

Answer 1

The Taylor series for `f(x)` centered at `a` is:

`f(x) = f(a)+f'(a)(x-a)+ (f''(a))/(2!)(x-a)^2+(f'''(a))/(3!)(x-a)^3+ * * *`

Explanation:

So the Taylor series for `f(x)` centered at `1` is:

`f(x) = f(1)+f'(1)(x-1)+ (f''(1))/(2!)(x-1)^2+(f'''(1))/(3!)(x-1)^3+ * * *`

Comparing the given function with the general expression for the Taylor series for `f` centered at `1`, we see that

`f(1) = 3`, and `f'(1) = -6`, and `f''(1) = 4`, and `f'''(1) = -5`, and finallya `f^((4))(x) = 1`.

So

`f''(1) = 4`, and `f'''(1) = -5`.

If you prefer , you could differentiate to get

`f(x)= 3- 6(x-1) + 4/(2!) (x-1)^2 - 5/(3!) (x-1)^3 + 1/(4!) (x-1)^4`

`f'(x) = -6+4(x-1)-5/(2!)(x-1)^2+1/(3!)(x-1)^3`

`f(x)= 4 - 5 (x-1) + 1/(2!) (x-1)^2`

So

`f''(1) = 4`, and `f'''(1) = -5`.