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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]

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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))]

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h

After more foiling and cancellations I got this:

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

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h

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

(-2c-h)

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

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

-2

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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?