How Can You Prove tan(x) = x Has Exactly One Solution in Each Interval?

[frankpupu]

Homework Statement

for the equation tanx=x in (npi-pi/2,npi+pi/2)for n is integers. show that it has precisely one solution.
without ploting the graph. use analysis

The Attempt at a Solution

i have try to assume two solutions then apply the rolles theorem but i cannot do in this way since assume two roots c1 and c2 then exists c s.t. f'(c)=o but f‘(x)=(tanx)^2>=0
can someone give me some helps

 

[Mark44]

Homework Statement

for the equation tanx=x in (npi-pi/2,npi+pi/2)for n is integers. show that it has precisely one solution.
without ploting the graph. use analysis

The Attempt at a Solution

i have try to assume two solutions then apply the rolles theorem but i cannot do in this way since assume two roots c1 and c2 then exists c s.t. f'(c)=o but f‘(x)=(tanx)^2>=0
can someone give me some helps

Look at the function f(x) = tan(x) - x.

Where are you getting f'(x) = tan²(x)?

 

[frankpupu]

Look at the function f(x) = tan(x) - x.

Where are you getting f'(x) = tan²(x)?

for f(x)=tanx-x
then in the range (npi-pi/2,npi+pi/2)
it is differentiable
then f'(x)=1+(tanx)^2-1=(tanx)^2

 

[Macch]

You mean (secx)^2 right?

Edit: Oh no, you're right.

 

[frankpupu]

You mean (secx)^2 right?

ok i get it thank you yes (secx)^2 then it will always greater than 0

 

[Ray Vickson]

Homework Statement

for the equation tanx=x in (npi-pi/2,npi+pi/2)for n is integers. show that it has precisely one solution.
without ploting the graph. use analysis

The Attempt at a Solution

i have try to assume two solutions then apply the rolles theorem but i cannot do in this way since assume two roots c1 and c2 then exists c s.t. f'(c)=o but f‘(x)=(tanx)^2>=0
can someone give me some helps

First you need to show (i) the equation has at least one solution. Then you need to show (ii) it cannot have more than one solution. Hint for (i): can you show the function tan x - x changes sign in the interval?

RGV

 

[Mark44]

ok i get it thank you yes (secx)^2 then it will always greater than 0

sec²(x) is always >= 1.

 

[frankpupu]

sec²(x) is always >= 1.

now i get confused again sec^2>=1 then sec^2-1 >=0 but we assume that we have 2 roots and use the rolles theorem then f'(c)=f'(d)=0 it means that it can have two root . how can i prove that only one root?

 

[Mark44]

Maybe I'm being dense, but I don't see why you think you need to use Rolle's Theorem. For that theorem, the conditions are that f(a) = f(b) = 0 (plus some other conditions), where a and b are the endpoints of some interval.

Furthermore, I don't think you have stated the problem correctly.

for the equation tanx=x in (npi-pi/2,npi+pi/2)for n is integers. show that it has precisely one solution.

In each of the intervals (n$\pi$ -$\pi$/2, n$\pi$ + $\pi$/2) there is a solution of the equation tan(x) = x, so in all of these intervals there are an infinite number of solutions. Do you need to show that in each interval there is exactly one solution?

If so, you can show this by looking at the derivative, and by determining whether the function itself changes sign, which is something Ray Vickson suggested a few posts back.