1. ## Prove the radical in the Equation

be cosx=x^2.

Prove that this equation has exactly two radical .
how i do it?

thanks for help

2. Originally Posted by ZOOZ
be cosx=x^2.
Prove that this equation has exactly two radical .
I think that you mean 'has exactly two roots'.
Look at the function $f(x)=x^2-\cos(x)$.
Show that it has two zeros in $[-1,1]$.
Then show its derivative is negative for numbers less than zero and positive for numbers greater than zero.

3. well cosx has two zeros in [-pi/2,pi/2] but i still no understand how to prove this function has exactly two roots maybe i need to use in Rolle's theorem can u Please show me how can i prove it?

thanks.

4. The zeros of cos(x) are irrelevant here. As Plato suggested, look at $x^2- cos(x)$ on the interval [-1, 1]. $1- cos(1)> 0$, $0^2- cos(0)< 0$, and $1^2- cos(1)> 0$.

The derivative of $cos(x)- x^2$ is $-sin(x)- 2x$. If you can show that derivative is negative for x> 0, as Plato suggested, then you know $cos(x)- x^2$ is "one-to-one" and has at most one zero for x>0. Similarly for x< 0.

5. 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?

6. Originally Posted by ZOOZ
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.
That post is practically unintelligible. Please practice using standard words, complete sentences and NO web-speak as ‘ur’.

$f(-1)=1-\cos(-1)>0,~f(0)=-1<0,~\&~f(1)>0$ that means it has a root between $-1~\&~0$ and a root between $0~\&~1$.
The derivative is negative for negative values of $x$ so there is one negative root.
Likewise there is one positive root.

7. Mean value theorem: If f is differentiable on [a, b] then there exist a number c, in [a, b], such that $\frac{f(b)- f(a)}{b- a}= f'(0)$. 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 $\frac{f(b)- f(a)}{b- a}= 0 = f'(c)$. 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 $cos(x)- x^2= 0$.

8. hi thank you for the asnwers but you have worng in 1 time you told to look at X^2-cosx on the interval [-1, 1] that ok.
BUT then you told The derivative of cos(x)-x^2.

so the derivative of cos(x)-x^2 is not like the derivative of X^2-cosx.

9. well ok i Understand how to Solve it now.

Thanks to you'r help.