# Cosh x > 1

1. Nov 23, 2009

### Jules18

1. The problem statement, all variables and given/known data
True or false?

Cosh x > 1 for all x

3. The attempt at a solution

the answer is true, and I'm not sure why. if h is a coefficient, then I'm pretty sure I can think of some values of x and h that would be < 1.
For instance, if h=1 and x=$$\pi$$/3

What does the question even mean?

2. Nov 23, 2009

### Staff: Mentor

cosh is short for hyperbolic cosine. h is not a parameter.

This function is defined as
$$cosh(x)~=~\frac{e^{x} + e^{-x}}{2}$$

Edit: no i in exponents

You might need to write the right side of the equation above as a Maclaurin series to show that cosh(x) >= 1 for all x.

Last edited: Nov 23, 2009
3. Nov 23, 2009

### jakncoke

cosh x = (Exp[x] + Exp[-x])/2

4. Nov 23, 2009

### Jules18

jakncoke, by Exp[x] do you mean ex where e is the constant that's approx. 2.718 ?

5. Nov 23, 2009

yes.