Help with showing one real solution

[native fiber]

Show that the equation has exactly one real solution: 2x+cosx=0

My process was
2x+cosx=0
x(2+cos)=0
ab=0 so
x=0

My question is am I allowed to factor x out of cosx like that or am i doing this problem wrong?

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[shadow wadi]

No, of course this is not okay

(To be clear, I understand that the problem is challenging, but please do NOT do that in a test)

[native fiber]

okay got it

which part of it is wrong? is it the way I presented it or are you not able to factor out x

[shadow wadi]

you are not able to factor x out

cos(x) is not something you just toy around with like that

it's a legitimate function of x like any other

[native fiber]

okay yeah I had doubts because cos itself doesnt have a value right

so it wouldnt make sense to factor it

[shadow wadi]

Indeed

But this problem is a bit technical

So I will give you some hints for you to solve it

And then I will go dry some clothes, come back and then I will answer your questions

[native fiber]

okay got it

thank you

[shadow wadi]

So an initial idea would be to plot those two functions

and see that their graph only has one intersection

Let me get desmos real quick

[native fiber]

I saw in desmos that

its just a vertical line

but if im doing it on paper to show it

[shadow wadi]

by two functions, I mean y = 2x and y = -cos(x)

since 2x + cos(x) = 0 <=> 2x = -cos(x)

So that's pretty good

we at least know the solution is not located in the positives

and it's easy to show too

So for now, let's assume that we successfully proved that there is no solution in R+ (you will have to prove it, but it's not the hardest part)

So I want you to answer the two following questions

1. why is there AT LEAST one solution in R-?

2. why is there ONLY one solution in R-?

You must use some theorems you've seen in analysis to answer

[native fiber]

like rolles or mean value?

[shadow wadi]

also don't forget that g(x) = 2x + cos(x) is differentiable

[shadow wadi]

yeah

this is not a problem that can be solved with brute force like you suggested

[native fiber]

got it let me work at this

you can do your laundry

[shadow wadi]

but using some theorems to get some hacks is good

[native fiber]

did you determine that the solution isn't located in the positives because of where the two separate functions intercept? im a bit lost on this part

[shadow wadi]

Case 1: x > 1/2, then clearly -cos(x) < 1 < 2x, so no solution found there
Case 2: 0 <= x <= 1/2, but since 1/2 is less than pi/2 then -cos(x) is negative, hence cos(x) < 2x

this is pretty straightforward

[shadow wadi]

In fact, with the same reasoning, it is clear that a solution is also not in (-infinity, -1/2]

[native fiber]

so by determining where its not

I can determine an interval

to use for one of the theorems?

[shadow wadi]

Well you could, yes

but it's pretty important that there is no positive solution, that's the main gist

so you can just focus on R-

or I guess (-1/2, 0)

[native fiber]

so could i use -1/2 and 0 as a and b and use the mean value theorem

[shadow wadi]

Does g have a zero in (-1/2, 0)? If yes, what theorem and what conditions ensure it

[native fiber]

hold on let me check my notes

[shadow wadi]

sure no problem

[native fiber]

would it be the intermediate value theorem

[shadow wadi]

with what value of a and b?

[native fiber]

-1/2 and 0 right because we determined that range

and the solution has to lie in between those values

[shadow wadi]

I mean you've gotta check right

g(0) = 2 * 0 + cos(0) = 1

but g(-1/2) = 2 * (-1/2) + cos(-1/2) = -1 + cos(1/2) by parity of cos

actually no you're right, that's pretty cool given that cos(1/2) < 1 hence cos(1/2) - 1 < 0

Ok, that's good

[shadow wadi]

And for the record, a solution has to lie between -1/2 and 0

We have yet to show that there is only one

but you did the first step already, so good job

Do you have any clue as for how to prove that there is only one solution?

[native fiber]

im thinking it through rn

its probably another theorem right

[shadow wadi]

Well, think about it

What if there were two solutions x1 =/= x2?

[native fiber]

it would change directions

and decrease

to cross the axis again

could i use the derivative like when calculating for critical points

[shadow wadi]

yeah you could

i mean, there's a theorem that even tells you about those hypothetical "critical points"

given two roots

[native fiber]

rolles theorem right

[shadow wadi]

yeah, you're correct

[native fiber]

because thats where when theres 2 roots

then f'(c) is 0 for them

[shadow wadi]

wdym "for them"?

be more precise

(I'm asking a lot of detail I know, but a lot of reasoning is expected from you in this exercise, so this is important)

[native fiber]

both the selected x values in the interval

[shadow wadi]

Well, let's say for convenience that x1 < x2

so where would c have to be?

[native fiber]

in between them both right

x1<c<x2

[shadow wadi]

included or excluded ?

[native fiber]

included i think

[shadow wadi]

That's correct

Since g(x1) = g(x2) = 0, according to Rolle's theorem, there exists c in [x1, x2] such that g'(c) = 0

Now try computing g'(c)

and that will be your cue

[native fiber]

thank you for walking me through this all

+close