Is Cosh x > 1 True for All Values of x?

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SUMMARY

The statement "Cosh x > 1 for all x" is true. The hyperbolic cosine function, defined as cosh(x) = (e^x + e^{-x}) / 2, yields values greater than or equal to 1 for all real numbers x. This can be demonstrated using the Maclaurin series expansion of the function, which confirms that cosh(x) >= 1 for all x. The confusion regarding coefficients and parameters in the discussion was clarified, emphasizing that h is not a variable in this context.

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Homework Statement


True or false?

Cosh x > 1 for all x

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?
 
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Jules18 said:

Homework Statement


True or false?

Cosh x > 1 for all x

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?
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:
cosh x = (Exp[x] + Exp[-x])/2
 
jakncoke, by Exp[x] do you mean ex where e is the constant that's approx. 2.718 ?
 
yes.
 

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