How do I find the roots/zeros of f(x)=tan(x)+(1/2)x ?

[forest coral]

I found that 0 is a root of f(x), but how do I find the other zeros?

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[onyx badger]

There are an infinite amount of solutions

$tan(x)+\frac{1}{2}x=0$ has one unique solution on the interval $]-\frac{\pi}{2} +n\pi,\frac{\pi}{2} +n\pi [$ for any natural number $n$

[opal arch]

Rotor 😑

[onyx badger]

But if you are restricted on one such interval then there is one solution and it is unique

[forest coral]

I checked on a graphing calculator, and this is what came out.

[onyx badger]

Yes, there are an infinite amount of solutions

[forest coral]

But on the interval [-2pi, 2pi] ?

[hasty pine]

do you know derivatives?

[forest coral]

Yes

[onyx badger]

You can’t define the function on this interval

x—>tan(x) diverges for x approaching pi/2 or -pi/2 mod(pi)

[forest coral]

But how would I find the roots without a graphing calculator?

[hasty pine]

do you need the EXACT roots?

or would approximations work?

you can use Newton's formula

[winged shoal]

Idk if it helps, but I found a formula that has 1 zero depending on the domain of the original function (tan(x)+x/2 = 0)

Formula: $arctan(x) + 2x + c\pi$, where $c$ is an integer

[opal arch]

clonesolopros

[forest coral]

Well it’s because I’m trying to solve this Calculus problem by using the first and second derivative tests. And I’m struggling to find the roots (interval endpoints) for the first and second derivatives.

The last image/graph is the correct answer, but I’m not able to get it.

[winged shoal]

Can you show your work? Even if its wrong

@forest coral

[forest coral]

Yes here

[winged shoal]

to find the roots, you first have to know that if a·b = 0, at least 1 of them has to be zero

[winged shoal]

Then to find the local minimum/maximum, y'=0
Then find y'' evaluated when y'=0

[forest coral]

I know. I’m struggling to find all of those points

[winged shoal]

Maybe the easiest way is to use this when c = [-2,-1, 0, 1, 2]

[rancid pine]

how are you sure it will have more roots then 1

[forest coral]

For the second derivative, there’s several changes in concavity, so there should be several roots there

[forest coral]

This is the correct answer, which I’m struggling to graph

It’s increasing on [-2pi, -pi], [0, pi] and decreasing on [-pi, 0], [pi, 2pi]

I can’t get the roots of -pi, and pi from the first derivative

[forest coral]

As to concavity points, the graph shows upward openings on a little before -pi to -pi/2, 0, pi/2, a little after pi to 2pi; and downward openings on -2pi to a little before -pi, -pi/2 to zero, pi/2 to a little after pi.

[winged shoal]

That is not correct

the roots of y are [-2pi, -pi, 0, pi, 2pi]

x = ±pi, and x = ±2pi, because sin(x) = 0

[hasty pine]

@forest coral

maybe it's not about using first derivative and second derivative tests

maybe it's just about using intermediate value theorem

[hot portal]

Well differentiate it and find point of inflexion maybe?I'm not sure but this should probably work