Thread: How to prove sinx+3-2x=0 has only one unique solution

I know sinx +3-2x=0 has only one unique solution that i found using Newtons algorithm in excel and from seeing the graph of sinx+3-2x=0 but i do not know how to prove this unique solution algebraically. Hopefully someone can help.

Thankyou

First define the function f(x)=sinx - 2x+3, this is continuous and differentiable function for all x.

Now, f(0)=sin0-2*0+3>0 and f(6)=sin(6)-12+3 ~ 1-12+3<0, from the Intermediate value theorem follows that there exist x_0 in (0,6) so that f(x_0)=0, hence we know that f(x) have one zero.


Now we will prove that above zero is unique!

We look at the derivative of f(x).

f'(x)=cosx-2<0 for all x. In other words the function f(x) is decreasing for every x, therefor the function is one-to-one, hence f(x)=0 only at x_0.

Originally Posted by nicolas123

I know sinx +3-2x=0 has only one unique solution that i found using Newtons algorithm in excel and from seeing the graph of sinx+3-2x=0

The derivative of is which is always negative.
The fact we have a decreasing function tells us what?

Thanks so sinx-2x=3=0 has one solution at the point 1.916 and then to prove it has one unique solution is that the first derivative cos(x)-2 decreases for every increasing value of x.
Cheers.

A decreasing function is one-to-one.

Originally Posted by nicolas123

I know sinx +3-2x=0 has only one unique solution that i found using Newtons algorithm in excel and from seeing the graph of sinx+3-2x=0 but i do not know how to prove this unique solution algebraically. Hopefully someone can help.

Thankyou

Alternatively,



The derivative of is whose maximum value is 1.

The slope of the line 2x-3 is 2, hence the sinewave can never cross over it a second time (the line slope would need to be <1 for this to be possible).