How do you show that the equation x-inverse tan(3x)=0 has a solution lying between x=1 and x=2? (Not sure if you need to use half-interval or newton's method??)

1 Answer
Apr 23, 2018

Since #f(x) = x-tan^-1(3x)# is a continuous function, all we need to show is that the signs of #f(1)# and #f(2)# are opposite.

Explanation:

The intermediate value theorem of real analysis states that if a continuous function, #f#, with an interval, #[a, b]#, as its domain, takes values #f(a)# and #f(b)# at each end of the interval, then it also takes any value between #f(a)# and #f(b)# at some point within the interval.

For this function, we can quickly check that

#f(1) = -0.249#

and

#f(2) = 0.594#

This means that #f(x)# takes every value between -0.249 and +0.594 at some value of #x# between 1 and 2. In particular, there is at least one value of #x# between 1 and 2 such that #f(x)=0#.

Note : The question asked for a demonstration that there is a root between 1 and 2 - it did not ask for the value of a root. What we have found is that #[1,2]# is what is called a bracketing interval - an interval which we are sure contains a root. Once we have a bracketing interval we can use any one of a large number of methods to find the root - one of them being the bisection method (I think that that is what is being referred to as the half-interval method here). One could, of course also use Newton-Raphson.