First of all, what you have is not an equation, but an expression.
Are you trying to solve for in the equation ? Or is it equal to some other number?
What you wrote is not an equation, it's an expression.
"bijection method" - well, you have to transform *stuff*=*stuff, stuff* in something like f(x)=f(y) and then prove that f is bijective (evidently, you chose a convenient function), so x=y. Then solve the new equation
Edit: o.o I was absolutly sure there is write "bijection". Sorry.
The first thing you need to do is find an interval in which you know there is a solution.
If we let , it's easy to see that f(0)= 0- 9(0)+ 1= 1> 0 and that f(1)= 1- 9+ 1= -7< 1. Since the value of this continous equation is postive at 0 and negative at 1, there must be a solution between them.
Now "bisection" means "cutting in half. Half way between 0 and 1 is, of course, 1/2. What is f(1/2)?
(Actually this is a surprisingly simple problem!)
thanks a lot .. nice
for a moment i was thinking about that equation ..
and the derivatives ... and the integrals .. and the fundamental theorem of calculus .. ( do i make any sense?)
then the differential equations ?
and after that the numerical methods ..
i wish all this stuff was much clearer ..