Given f(x) = 3 - 5x + x^3, how can I show that f(x) = 0 has a root (x=a) in the interval 1<a<2?

In plain english, we need to show that there is a value of x, which we call "a", in the interval 1 < a < 2 where f(a)=0. To prove this we start by letting x = 1: f(1) = 3 - 5(1) + 13 = -1. We now let x = 2: f(2) = 3 - 5(2) + 23 = 1. Since there is a change of sign of the value of f(x) in the interval of 1 < x < 2, then there must be a value of x = a where f(a) is zero. Therefore, the function f(x) = 0 has  a root (x = a) in the interval 1 < a < 2.

Answered by Giorgos P. Maths tutor

6923 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Integrate Sin(x)Cos(x)dx.


A curve is defined by the parametric equations x=(t-1)^3, y=3t-8/(t^2), t is not equal to zero. Find dy/dx in terms of t.


Differentiate the function f(x)=2xsin3x


Is the trapezium rule an exact method of integration?


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