Prove that the function f(x) = x^2 - cos x satisfies f(x)=0 for precisely two numbers x.?

Question

Prove the same for the function f(x) = 2x^2 - x sin x - (cosx)^2.
(Some preliminary estimates will be useful to restrict the possible location of the zeros of f.)

Please show your work. thank-you!

Use Rolle's Theorem to prove the functions.

Answer 1

well okay:
since f(x) = 0
so x^2 = cosx
and x = cosx^1/2
and so you can find the values of x

Answer 2

x^2 is always positve.

the minimum of -cos(x) is -1.

thus the addition of -cos(x) to x^2 can at most drop the graph below the x-axis by -1.

now, when x = pi/2 cos(x) = 0 ans f(x) = (pi/2)^2 = 4.935 which is already greater then 1. thus for all x > pi/2 f(x) is positive.

lastly since -cos(x) is strictly increasing on the interval x = 0 to x = pi/2 (as is x^2) there exists 1 and only one x for which f(x) = 0.

the same holds true for the interval x = 0 to x = -pi/2.

Answer 3

x^2 - cos x = 0

For x = 0

f(x) = 0^2 - 1 = -1

for x=pi/2

f(x) = (pi/2)^2 - 0 > 0

Because f is continuous there's an x between 0 and pi/2 such that f(x)=0

Let us derive:
f'(x) = 2x + sin(x) [Remark: sin(x) is positive in (0, pi/2), and in [pi, inf) thus f(x) is monotonously increaing within this interval.]

f(x) = x^2 - cos x >= x^2 - 1 >= (pi/2)^2 - 1 > 0

Thus there is only 1 positive x for each f(x) = 0

The same way, we prove that there must be an x between -pi/2 and 0 such that f(x) = 0, and thus there's only one such negative x.

Answer 4

f(x)-cos(x)=f(-x) - cos(-x) that's wy this equation has 2 or 4 or 8... zeros.
but it can't has more then 2 zeros, becose x^2 always grows(when x>0) and when cos(x)=0 (x=Pi/2), x^2 is greater then 1. That's wy them newer cross each other, when x>Pi/2, and we have 1 root when (x>0)and(x
When x<0 we also have 1 root.

that's wy we have 2 roots simmetrically.

I can't explain more detail, becose my english is't wery well.

Answer 5

f(-pi/2)>0
f(0)<0
f(pi/2)>0
and f is increasing for x>0 and decreasing for x<0.
Also, outside [-pi/2,pi/2] f(x) >1
P.S.(edited): Naturally f is continue, and this is essential for this reasoning

Answer 6

Yeah...I think that is not correct.