Answer:

Step-by-step explanation:
If
, then
. It follows that
![\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20%5C%5C%5Cfrac%7Bg%28x%2Bh%29-g%28x%29%7D%7Bh%7D%20%26%3D%20%5Cfrac%7B1%7D%7Bh%7D%20%5Ccdot%20%5Bg%28x%2Bh%29%20-%20g%28x%29%5D%20%5C%5C%26%3D%20%5Cfrac%7B1%7D%7Bh%7D%20%5Cleft%28%20%5Cfrac%7B1%7D%7Bx%2Bh%7D%20-%20%5Cfrac%7B1%7D%7Bx%7D%20%5Cright%29%5Cend%7Baligned%7D)
Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.

Answer:
-15
Step-by-step explanation:
as i said if the signs are different the result is negative
Answer:
I think it would be 2 infinity