If you view graphically the two equations
f1(x)=sin(x) ....(1)
f2(x)=2x-1 ....(2)
it will be obvious that there is only one real root.
To prove that there is only one real root, you could:
1. determine the interval I of x on which f2(x) is between -1 and 1.
2. find the root x0 where f1(x)=f2(x).
3. analyse the sign of the function f2(x)-f1(x) on the interval I on each side of x0.
show that the equation 2x-1-sinx=0 has exactly one real root.
sin x = 2x-1
2x = sin x - 1
range of sin x is -1 to + 1
but how do you know that 2x is between -2 and 0 and so x is between -1 and 0
3 answers
Think graphically. The curve of sinx -1 oscillates between -2 and 0, touching zero at integer multiples of x = pi. It is -2 at x = 0.
The curve y = 2x can only intersect sinx -1 at one point, somewhere between x = -pi and 0. For x > 0 and x < -pi, it is too large to ever intersect sinx -1 again.
The curve y = 2x can only intersect sinx -1 at one point, somewhere between x = -pi and 0. For x > 0 and x < -pi, it is too large to ever intersect sinx -1 again.
The curve of sinx -1 oscillates between -2 and 0, touching zero at ODD integer multiples of x = pi. The argument remains the same for the reason there is only one root.
1 answer