If the linearization of y= cube root of x at x=64 is used to approximate cube root of 66, what is the percentage error?

Question

a) .01%
b).04%
c).4%
d)1%
e)4%

Answer 1

For the calculation of the error term ε, see response to:

http://www.jiskha.com/display.cgi?id=1291926261

In the given case,
h=2
a=64
f(a)=4
f(a)+hf'(a)=4.041667

f(x)=x^(1/3)=
f'(x)=1/(3*x^(2/3))
f"(x)=-2/(9*x^(5/3))
and

epsilon(x,h)=h²f"(x)/2!
=-2h²/(9*x^(5/3))
=-0.000434

actual error
=f(66)-4.041667
=-0.000427

Please double-check my calculations and make your choice for the correct answer.