Finding Horizontal Tangents and Derivatives

[dee11]

Can anybody please help me, I need to know if I'm headed in the right direction, I haven't had math in a while and getting this together is proving to be difficult.

I may be totally wrong but here is the work I have done. This is my first time on here so, if I'm not showing enough work, I apologize. I feel like the village idiot in this class.

Show that if g(u)= u - 1/u, then dg/du (c) = ((c^2) + 1 )/ (c^2)

I'm assuming that you solve for the derivative from any point c.

[(c+h)^2 + 1)/ (c+h)^2)] - [(c^2 + 1) /c^2]
----------------------------------------------
h

Then I foiled and combined made the denominators the same to combine the top fractions

[(c^2 (c^2 + 2ch + h^2 + 1))/(c ^2 (c^2 + 2ch + h^2))] - [(c^2 + 1)(c^2 + 2ch + h^2)/(c^2(c^2 + 2ch + h^2))]
-----------------------------------------------------------------------------------------
h

After more foiling and cancellations I got this:

(-2ch - h^2)/ (c^2(c^2 + 2ch + h^2))
---------------------------------------
h

Then I multiplied by the reprical of h and factored out the -h afterwards and got...

(-2c-h)
-----------------------
(c^2(c + 2ch + h^2))

I then took the limit of that as h--> 0 and got


-2
------
c^2

This doesn't seem right.




Also, one simple question to find all values of x where the curve y = x^5 - 15x^3 + 251 has a horizontal tangent line.

I think that means that where the slope is equal to 0.

So if I set this equal to 0, would that work?

 

[hotvette]

Welcome to the forums.

Re the first problem, you've applied the fundamental theorem of differentiation to dg/du instead of g.

Re the 2nd problem, you are correct about the answer being where the slope is zero. But the second part, if by 'this' you mean the equation provided, that will just tell you where y = 0. Hint: what calculus concept has to do with slope?