The zeros of cos(x) are irrelevant here. As Plato suggested, look at on the interval [-1, 1]. , , and .
The derivative of is . If you can show that derivative is negative for x> 0, as Plato suggested, then you know is "one-to-one" and has at most one zero for x>0. Similarly for x< 0.
well ok for f'=-sin(x)-2x the derivative is negative for x>0
and for f'=-sinx-2x the derivative is Positive for x<0
this i Understand but when u told "then you know cos(x)-x^2 is "one-to-one" and has at most one zero for x>0. Similarly for x< 0."
for what i need all the part A of ur answer i just can use the part with the derivative no?
That post is practically unintelligible. Please practice using standard words, complete sentences and NO web-speak as ‘ur’.
that means it has a root between and a root between .
The derivative is negative for negative values of so there is one negative root.
Likewise there is one positive root.
Mean value theorem: If f is differentiable on [a, b] then there exist a number c, in [a, b], such that . It there were more than one root of the equation f(x)= 0 we could take a and b to be those roots and have . That is, if there is more than one root, there would have to be a point beteen them where f'(c)= 0. But f'(x) here is 0 only at x= 0. There cannot be two positive or two negative value of x such that .