find the critical points of f on the domain or given interval
$$(e ^{x}+e ^{-x})/2$$

Question

find the critical points of f on the domain or given interval
$$(e ^{x}+e ^{-x})/2$$

zehanz · Answer

You'll have to look for the derivative, and solve:

((e^x+e^-x)/2)' = 0

anonymous · Answer

the derivative is the exact same thing as the original in this case

anonymous · Answer

but I'm not sure how they got 0 as the answer

zehanz · Answer

It's not exactly the same: (e^x)' = e^x, but (e^-x)' = -e^-x

zehanz · Answer

So, (the factor two is not important) you end up with e^x - e^-x=0
Or: e^x=e^-x, which has only 0 as a solution.

anonymous · Answer

oh wow stupid mistake, my fault. sorry about that lol

zehanz · Answer

No problem, happens to all of us from time to time...

anonymous · Answer

A critical point is where the graph of the function turns. This is where the first derivative is zero.
So for, 
$$y =\frac{ ( e^x +e ^{-x}) }{ 2 }$$
So,
$$\frac{ d }{ dx } (\frac{ ( e^x +e ^{-x}) }{ 2 }) = \frac{ 1 }{ 2 } (e^x - e^{-x} )$$
When does this equal zero? Well, it equals zero when, 
$$\frac{ 1 }{ 2 } (e^{x} - e^{-x}) = 0 $$
$$e^{x} - e^{-x}=0$$
$$e^{x}=e^{-x}$$
Taking the natural logarithm on both sides, we have 
$$x = -x$$
And so,
$$x=0$$