# Hyperbolic Functions

• Oct 25th 2009, 09:47 PM
xxlvh
Hyperbolic Functions
Consider $f(x) = asinhx + bcoshx$ where a and b are real numbers.
For what values of a and b does f have an extremum at x = 0?
For what values of a and b does f have an extremum at x = c?

I started out by differentiating:
$f'(x) = acoshx + bsinhx$
At extremum f'(x) = 0. So setting f'(x) = 0 and x = 0,
$0 = a*1 + b*0$
? Doesn't seem to be correct.
I don't know where to go from here to solve this problem. If anyone could help me out, would be greatly appreciated, I don't understand the overall concept of hyperbolic functions very well. Thanks!
• Oct 25th 2009, 10:10 PM
scorpion007
Set f'(x) = 0 and solve for a, b given x.

x=0: $a(1)+b(0)=0 \implies a=0, b \in R$

At a general point x=c:

$a\cosh{c} + b\sinh{c}=0$

$\Rightarrow a\cosh c = -b\sinh c$

$\Rightarrow a = -b\frac{\sinh c}{\cosh c} = -b\tanh c, ~b\in R$
• Oct 25th 2009, 10:22 PM
xxlvh
Thanks a bunch! At least I was on the right track with this question, that's a relief.
Would you mind explaining the last point a bit more though?
$a = -btanhc$ and how that goes to b is any real number.
• Oct 25th 2009, 10:25 PM
scorpion007
$\sinh{x}/\cosh{x}=\tanh x$.

And $b\in R$ means b is any real number. (I just solved for a in terms of b, since there are infinitely many solutions)

The solution to the equation $f'(c) = 0$ can be expressed as a set $S=\{ (a,b) \mid a=-b\tanh{c}, ~b\in\mathbb{R}\}$.

I hope that's right.
• Oct 27th 2009, 05:04 PM
xxlvh
I am unsure if the term "extremum" refers to local or abolute min, I don't know if it applies, however is it sufficient to only set the derivative to equal zero for extremum or will the second derivative test need to be used to verify that these are not points of inflection?
• Oct 27th 2009, 07:50 PM
scorpion007
You are correct! You must ensure that the second derivative is not 0, otherwise you don't (necessarily) have an extremum.

An extremum is any local or global max or min.