Finding how many solutions a function has

Asked Jun 08 '18 at 07:11

Active Jun 08 '18 at 08:24

Viewed 128 times

I have the function

$$f (x) = \cos x - \frac{x^2}{100}$$

$$f'(x) = -\sin x \, -\frac{x}{50}$$

$$f''(x) = -\cos x - \frac{1}{50}$$

and I want to find out how many roots it has, I have tried using calculus and differentiating it. I have tried using Newton's method, and then realized it's not that helpful if I want to find HOW MANY instead of where the roots are.

edited Jun 08 '18 at 08:24

Richard Ambler

1,665

asked Jun 08 '18 at 07:11

Guysudai1

You can use the mean value theorem to find the number of roots – The Integrator Jun 08 '18 at 07:14
@TheIntegrator thanks. Ill research what that is. I've seen you in one of my other posts yesterday, you sure are helpful – Guysudai1 Jun 08 '18 at 07:17
2

Happy to help. here are some links which might better explain it 1, 2 – The Integrator Jun 08 '18 at 07:19
2

You can at least reduce the search space by noticing that any roots that exist must lie in $-10\le x\le 10$, since $-1\le \cos x \le 1$... – Richard Ambler Jun 08 '18 at 08:25

Finding how many solutions a function has

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