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

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- Nov 21st 2010, 11:58 AMnicolas123How 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 - Nov 21st 2010, 12:01 PMAlso sprach Zarathustra
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. - Nov 21st 2010, 12:05 PMPlato
- Nov 21st 2010, 12:33 PMnicolas123
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. - Nov 21st 2010, 12:39 PMPlato
A decreasing function is one-to-one.

- Nov 21st 2010, 01:07 PMArchie Meade
Alternatively,

$\displaystyle sinx=2x-3$

The derivative of $\displaystyle sinx$ is $\displaystyle cosx,$ 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).