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Domain/Range of Cosh(x)

Right. This is beginning to annoy me.

In the P5 textbook, it has a diagram of the graph of y = cosh(x), in which it is defined over both positive and negative x.

Later it says:

For the function cosh(x), you need to take the domain x>0, so it is a one-one function. Then the inverse function arcosh(x) is defined for the domain x>1 and range arcosh(x)>0.

I realise that to have an inverse, a function must be one-one. But do we always want to restrict the domain of cosh(x) ?

Since cosh(x) is even, when for example solving quadratics in cosh(x) (for x), I'm often left with +/- something and the back of the book sometimes includes both or usually just the positive answer - but I can never figure out why it's one or the other.

Any help would be much appreciated. :smile:

Reply 1

AlphaNumeric

10

Typically inverses are defined so you get positive values back, it's just convention. If you're dealing with physical quantities like time or length, then you'll want a positive value most of the time too. If you don't restrict the domain, you end up with multivaluedness which is an annoyance to many people :p:

Reply 2

meef cheese

OP

10

So you reckon if I come out with something like x = +/- ln3, I just accept +ln3?

Reply 3

dvs

10

No, cosh(+/- ln3) is well-defined; arcosh(-ln3) isn't.

Reply 4

meef cheese

OP

10

dvs

No, cosh(+/- ln3) is well-defined; arcosh(-ln3) isn't.

You see, that makes sense to me but I don't understand the following question:

Solve 3sinh²(x) - 2cosh(x) - 2 = 0

3(cosh²(x) - 1) - 2cosh(x) - 2 = 0
=> 3cosh²(x) - 2cosh(x) - 5 = 0
=>(3coshx - 5)(coshx + 1) = 0
=> coshx = 5/3 and coshx = -1 (no solutions)
=> x = arcosh(5/3)

From a sketch of the graph of y = coshx, it's clear that both +/- ln3 will give the correct answer.

But according to the definition of arcosh given earlier, you only get +ln3

arcosh x = ln[x + sqrt(x² - 1)] (taking only the positive square root)
= ln[5/3 + sqrt(25/9 - 9/9)
= ln[5/3 + sqrt 16/9]
= ln(9/3)
= ln 3

Yet in the answers it gives =/-ln3, hence my confusion.

Maybe, this is just a question of what the A-level markers are looking for. Does anyone know?

Reply 5

Gaz031

15

coshx=y => x=+/- arcosh(y) so you should get +/- ln3.

Reply 6

meef cheese

OP

10

I'm sorry, I still don't get this.

If I use arcosh(x) = ln[x + sqrt(x² - 1)] I don't get the negative answer.

Am I using the logarithmic form in the wrong instance or something?

Reply 7

Gaz031

15

e-unit

I'm sorry, I still don't get this.

If I use arcosh(x) = ln[x + sqrt(x² - 1)] I don't get the negative answer.

Am I using the logarithmic form in the wrong instance or something?

In order for coshx to have an inverse, you specify x>=0. So coshx=y => x=arcoshy for x>=0.
You then note that cosh(x)=cosh(-x) so if x satisfies coshx=y then it follows that (-x) also satisfies cosh(-x)=y.
If you're simply solving the equation, both positive and negative values can be solutions.

Reply 8

7589200

18

but mark schemes always seem to take the postive solution...

Reply 9

meef cheese

OP

10

Vazzyb

but mark schemes always seem to take the postive solution...

Exactly ... well, I thought they did until the answers at the back of the book started including negative ones too.

Once I start doing past papers, I'll probably just stick to whatever's in the markscheme.

Reply 10

dvs

10

Can you give examples where only a positive answer is included?

Reply 11

dvs

10

e-unit

=> coshx = 5/3
=> x = arcosh(5/3)

I think this is your mistake. It does follow that x=arcosh(5/3) is a solution, but it doesn't follow that it's the only one. In fact the solutions are x = +/- arcosh(5/3) = +/- ln(3).

Remember that cosh is even, so cosh(-x) = cosh(x) for all x.