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Having trouble getting the answers to these questions...

If f and g are differentiable increasing functions and g(x) is never equal to 0, then the function h(x) = f(x)/g(x) is also a differentiable increasing function.

If a function is period with period c, then so is its derivative.

If C(q) represents the cost of producing a quantity q of good, then C'(0) represents the fixed costs.

Thanks!

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    To give a start, let $g(x)=e^{2x}$2011-11-15

2 Answers 2

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For the first, it is true that $h$ is differentiable. But if the conclusion is true, then take an example in which neither $f$ nor $g$ are ever equal to $0$. Then the conclusion would be that $f/g$ and $g/f$ are both differentiable increasing. But if $f/g$ is increasing, then $g/f$ has to be decreasing! So the conclusion cannot be always true.

For the second, consider the derivative as a limit, and use the fact that $f(a)=f(a+c)$.

The third one is false. C'(0) represents the marginal cost to start producing something; i.e., how much more it will cost you to produce 1 item than it costs you to produce no items. That's not the fixed costs.

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1) False. $f/g$ would indeed be differentiable; but, not neccesarily increasing. For example, take $f(x)=e^x+1,{\rm\ and\ } g(x)=e^{2x}+1.$ Then $\bigl(f/g\bigr)(0)=1$ but $\bigl(f/g\bigr)(\ln (2)) =3/5$ (or look at Arturo's nicer argument above).

2) True. If $f$ has period $P$ and is differentiable, write the derivative as f'(c)=\lim_{h\rightarrow0}{f(c+h)-f(c)\over h} =\lim_{h\rightarrow0}{f(c+P+h)-f(c+P)\over h}=f'(c+P).

3) False. The fixed cost is $C(0)$.