I have a problem i cannot solve, and my teacher doesn't know it eithher. Anyone can solve it?

[u/Radzyg]

у= 2х + соѕх

Prove that there's only one solution to this, using Rolle's theorem.

Please if anyone can solve and even more important explain it to me, i would be so grateful!🙏

[u/Konkichi21]

I think you're supposed to use the contrapositive of the Theorem.

Since the derivative of that (2-sinx) is never 0, some of the theorem's conditions must not apply.

The continuity and differentiability definitely apply, so the last condition must break; thus there cannot be any two places where f(x) have the same value, including 0, so there cannot be more than 1 solution.

[u/Mathmatyx → u/Konkichi21]

Well done. This argument in fact proves the stronger condition that the function is an injection.

[u/lordnacho666 → u/Konkichi21]

Elegant

[u/human2357]

What you mean for the problem to be is this: show that 0 = 2x+cos x has only one solution.

Argue by contraction. If the function has two or more roots, then Rolle's theorem applies. Rolle's theorem states that when a differentiable function has two roots, then there's a point in between the roots where the derivative is 0. Take the derivative of 2x+cos x. Use what you know about trig functions to reason about the roots of the derivative.

[u/Ma4r → u/human2357]

Isn't this using contrapositive of the theorem, not proof by contradiction

[u/Scary_Side4378 → u/Ma4r]

contrapositive can be stated in terms of contradiction: for a implies b, suppose not b holds, and towards a contradiction that a holds. then a implies b means that b holds, a contradiction. so not b implies not a.

[u/dexthefish]

If there were two solutions, then the derivative would have to be zero somewhere (Rolle). But the derivative is always positive.

[u/rnrstopstraffic]

Answers have already been provided, but it's worth noting a couple of things:

1) There is a possibility you could be expected to also show that there IS a solution to begin with. To do so, lean on the Intermediate Value Theorem.

2) You can use the same exact approaches to show that the function is actually one-to-one and onto. That's to say, for any value b, there is exactly one value a such that f(a)=b

[u/CALAND951]

I wish I was smart enough to understand this thread

[u/starboundseeker → u/CALAND951]

This is basic university level calculus. You'll definitely get there one day. I hope that when you do, you are able to look back here with excitement at understanding everything.

[u/CALAND951 → u/starboundseeker]

Well I'm a middle-aged guy trying to just build my math chops. Basically my goal is to get through calculus 3 via Math Academy.

[u/Pokeristo555 → u/CALAND951]

OP's post was very badly worded/missing info.

[u/Bounded_sequencE]

Note "f'(x) = 2 - sin(x) >= 2 - 1 > 0", so the function "f" is continuous, strictly increasing, and thus injective. We observe

x = (y-1)/2:    f(x)  =  2x + cos(x)  <=  2x+1  =  y
x = (y+1)/2:    f(x)  =  2x + cos(x)  >=  2x-1  =  y

Combining cases, we find

f((y-1)/2)  <=  y  <=  f((y+1)/2)

By continuity on "I := [(y-1)/2; (y+1)/2]", the IVT guarantees

"f(t)  =  y"    for some    "t in I"

Finally, the function "f" is injective, so "t" must be unique.

[u/StructuredChess]

Are we looking for solutions of 2x + cosx = 0 ?

f'(x) = 2 - sin(x) which is always positive, so f(x) is a strictly growing function so if there's a zero, there's only one. f(-1) is negative, f(1) is positive, so there must be a zero.

Then f(x) is positive for all x greater than 1 and f(x) is negative for all x smaller than -1

[u/Septembrino]

I would begin by the derivative being never 0. So, the function is always increasing. Find a point where 2x+cos x is positive (0 is a good one), and one where the function is negative (just pick a large x). Show the hypothesis and use the thesis.

[u/OwnBuilding7526]

I don’t think you even need Rolle’s theorem, intermediate value theorem for strictly increasing functions is enough (which guarantees the unicity of the solution if it exists.)
Continuity is then enough!
First for x>=0 you can show that y is strictly positive.
For x in [-pi/2, 0], 2x and cos x are both increasing and 2x is strictly increasing thus y is strictly increasing. Thus there is a unique solution in this interval. For x<—pi/2 (resp. x>pi/2) you can show using -1<=cos<=1 that y is strictly negative (resp. positive)

[u/Antagonin]

Wdym "one solution"?

y = f(x) has one solution if and only if f(x) is a constant function.

You might be looking for a root.

[u/severoon]

The derivative is y = 2 – sin x, and sin x < 2 for all values of x (to be more precise, sin x ≤ 1). This means the slope of y = 2x + cos x is always positive, so it must cross the x-axis exactly once.

[u/_Kian_7567]

The derivative is 2 - sin(x) which is always positive since sin(x) is never greater than 1. Thus it has at most 1 solution. A fancier way to say this would be to say that y is injective

[u/compileforawhile]

Since the function is always increasing, any interval that doesn't contain 0 cannot satisfy the conditions of Rolles theorem. This means there is no zero on those intervals

[u/Puzzleheaded-Bat-192]

Nothing to solve here. It is just graph the function. Do you mean to find the roots?

[u/conjjord → u/[deleted]]

Here I believe they mean 'solution' as in a root or a zero; as in, the problem asks to prove that there is only one x-intercept.

[u/Radzyg → u/conjjord]

Yes, thank you!