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

Answered by Sanjith H. Maths tutor

2412 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find the equation of the line tangential to the function f(x) = x^2+ 1/ (x+3) + 1/(x^4) at x =2


A curve has the equation 2x^2 + xy - y^2 +18 = 0. (1) Find the coordinates of the points where the tangent to the curve is parallel to the x-axis.


Prove by induction that, for n ∈ Z⁺ , [3 , -2 ; 2 , -1]ⁿ = [2n+1 , -2n ; 2n , 1-2n]


Find values of x in the interval 0<x<360 degrees. For which 5sin^2(x) + 5 sin(x) +4 cos^2(x)=0


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences