Use Taylor series to approximate a number.

[Sabricd]

Hello,
I need help with this problem. I need to find the first three terms of the Taylor series for the function f(x)= (1 + x)^(1/3) to get an estimate for 1.06^(1/3).

Hence I did:
f(x)= (1 + x)^(1/3)
f'(x)= (1/3)(1 + x)^(-2/3)
f''(x)= (-2/9)(1 + x)^(-5/3)
f(a) + f'(x)/1! * (x - a) + f"(a)/2! * (x - a)^2

Now, I'm kind of stuck...Could I please have a hint on how to finish the problem.
Thank you,
-Sabrina

 

[Mark44]

Your Taylor's series should have x terms in it. (You're really getting the Maclaurin series for your function, since you are expanding about x = 0.)

You still need to evaluate the function and its two derivatives at x = 0 in order to approximate f(.06).

$$f(x) \approx f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2$$

 

[Sabricd]

Thank you so much Mark! :)