If y^3 = 8.08, approximate y.

Firstly, recognize that 2^3 is 8, so y must be close to 8.It will be helpful to then write y^3 as y^3= 8 + 0.08We can then factorize out 8.y^3=8(1+0.01)If we try and take the cube root of this expression.y=2(1+0.01)^(1/3)
We recognize this is a binomial expansion, if we label x as 0.01 we can see a more familiar form y=2(1+x)^1/3
Expanding this and truncating the expansion for the first order term, we are left with y = 2 + 2x/3
Substituting in, x=0.01 we get y being roughly equal to 2.01

SH
Answered by Sanjith H. Maths tutor

2915 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do I remember the coefficients of a Taylor expansion?


What is the derivative with respect to x of the function f(x)=1+x^3+ln(x), x>0 ?


Find the values of x where the curve y = 8 -4x-2x^2 crosses the x-axis.


Let w, z be complex numbers. Show that |wz|=|w||z|, and using the fact that x=|x|e^{arg(x)i}, show further that arg(wz)=arg(w)+arg(z) where |.| is the absolute value and arg(.) is the angle (in polar coordinates). Hence, find all solutions to x^n=1 .


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning