By using the correct theorems show that 3 x - 4 = cos has at least one real root and at most one real root

Question

Consider the equation 3 x − 4 = cosx. Please show (by referring to the correct Theorems and verifying that their assumptions are satsified!) that the equation has

(a) at least one real root.

(b) at most one real root.

Answer 1

First, you need to show that both both the right side and the left side are continuous everywhere.

Then you would show that 3x-1 can only between -1 and 1 when 1≤x≤5/3 i.e since cos x takes only values between -1 and 1 ,if there is a root, it must be in the interval where the left side is in that interval. There is no theorem for that!

Then I would show that 3x-1-cos x is less than 0 at -1 and greater than 0 at 5/3

This means that the difference has a zero between x=1 and x=5/3, i.e. the equation has a root in that interval..

There is a theorem which states precisely that (it is related to but not the same as the Intermediate Value Theorem.)

To prove there is only one root I would show that the derivative of the two sides are opposite in sign on the relevant interval and no turning points...I don't know what theorem to quote for that But there may be a better way which I do not see at the moment.