be cosx=x^2.
Prove that this equation has exactly two radical .
how i do it?
thanks for help
The zeros of cos(x) are irrelevant here. As Plato suggested, look at $\displaystyle x^2- cos(x)$ on the interval [-1, 1]. $\displaystyle 1- cos(1)> 0$, $\displaystyle 0^2- cos(0)< 0$, and $\displaystyle 1^2- cos(1)> 0$.
The derivative of $\displaystyle cos(x)- x^2$ is $\displaystyle -sin(x)- 2x$. If you can show that derivative is negative for x> 0, as Plato suggested, then you know $\displaystyle cos(x)- x^2$ 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’.
$\displaystyle f(-1)=1-\cos(-1)>0,~f(0)=-1<0,~\&~f(1)>0$ that means it has a root between $\displaystyle -1~\&~0$ and a root between $\displaystyle 0~\&~1$.
The derivative is negative for negative values of $\displaystyle x$ 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 $\displaystyle \frac{f(b)- f(a)}{b- a}= f'(0)$. 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 $\displaystyle \frac{f(b)- f(a)}{b- a}= 0 = f'(c)$. 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 $\displaystyle cos(x)- x^2= 0$.