Linear Approximation of \sqrt[3]{27.02} using f(x)+f'(x)(x-a) method

[forestmine]

Homework Statement

Find the linear approximation of $\sqrt[3]{27.02}$

Homework Equations

f(x)+f'(x)(x-a)

The Attempt at a Solution

So what I did was work with$\sqrt[3]{27}$ since that's an easily known value. So my f(x)=x$^{1/3}$ and my f'(x)=1/3x$^{-2/3}$. From there, I worked f(27) = 3, and f'(27)=1/27.

Then I used the above equation. 3+1/27(x-27). For my value of x, I used 27.02, and got .0004.

Something tells me I'm doing something incorrectly, though...

Thanks for the help!

 

[hotvette]

Are you saying 3+1/27(x-27) = .0004? It has to larger than 3.

 

[forestmine]

My mistake -- calculated that incorrectly. I actually get 3.00074. Was my method correct in that case?

 

[hotvette]

My mistake -- calculated that incorrectly. I actually get 3.00074. Was my method correct in that case?

Your method looks ok to me. A good sanity check is to calculate the cube root of 27.02 and compare to the approximation.

 

[forestmine]

Thank goodness for sanity checks. Thanks a lot!