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
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.
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).