f(x) = sqrt( (e^x) - 1 ): find the domain, find 1st and 2nd derivatives, and test for concavity

[boywithmathproblems]

Hi!

This problem is giving me a headache. I'm practicing for an exam and this one is from one of the old exams. Anyway,

We had to find the domain, first and second derivative and where the function is concave and convex.

The problem for me starts at the second derivative part. The expression just seems to complicated, I'm probably missing some basic knowledge here or something, but I don't even know how to simplify it hence I don't know how to check the concavity and all that. The picture is where I get stuck, and whatever I try I don't get a nice result. I try rationalizing it, I tried the common denominator. I tried coming back to it after a day, but I guess I keep doing the same mistakes.

If you read this, thank you for your attention!
If I should post this in Calculus let me know, but it seems like an algebraic problem to me.

 

[boywithmathproblems]

Hi!

[math]f(x) = \sqrt{\smash[b]{e^x-1}}[/math]
I have to find the domain, first and second derivative and test for concavity

I'm getting stuck on the second derivative part, whatever I do after the step in the attached pic I just get a unusable mess of a function I don't know how to use to test for concavity.

 

[stapel]

[math]f(x) = \sqrt{\smash[b]{e^x-1}}[/math]
I have to find the domain, first and second derivative and test for concavity

I'm getting stuck on the second derivative part, whatever I do after the step in the attached pic I just get a unusable mess of a function I don't know how to use to test for concavity.View attachment 37422

I get the same result. Now try simplifying the second half of the expression by breaking the one fraction into two:

[imath]\qquad \dfrac{1}{2}\left(\dfrac{e^x\, sqrt{e^x - 1\,}}{e^x - 1} - \dfrac{e^x \cdot e^x}{2\sqrt{e^x - 1\,}(e^x - 1)}\right)[/imath]

[imath]\qquad \dfrac{1}{2}\left(\dfrac{2e^x\, \sqrt{e^x - 1\,}\sqrt{e^x - 1\,}}{(e^x - 1)\sqrt{e^x - 1\,}} - \dfrac{e^x \cdot e^x}{2\sqrt{e^x - 1\,}(e^x - 1)}\right)[/imath]

See where that might lead.

 

[blamocur]

Multiplying both numerator and denominator by [imath]2\sqrt{e^x-1}[/imath] worked better for me.