# Equation solution problem

1. Feb 20, 2012

### frankpupu

1. The problem statement, all variables and given/known data
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

3. 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

2. Feb 20, 2012

### Staff: Mentor

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

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

3. Feb 20, 2012

### frankpupu

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

4. Feb 20, 2012

### Macch

You mean (secx)^2 right?

Edit: Oh no, you're right.

5. Feb 20, 2012

### frankpupu

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

6. Feb 20, 2012

### Ray Vickson

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

7. Feb 20, 2012

### Staff: Mentor

sec2(x) is always >= 1.

8. Feb 21, 2012

### frankpupu

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?

9. Feb 21, 2012

### Staff: Mentor

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.
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.