Respuesta :
Answer:
[tex]\dfrac{-1}{x(x+h)}, h\ne 0[/tex]
Step-by-step explanation:
If [tex]g(x) = \dfrac{1}{x}[/tex], then [tex]g(x+h) = \dfrac{1}{x+h}[/tex]. It follows that
[tex]\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}[/tex]
Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.
[tex]\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}[/tex]
Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.
[tex]= \dfrac{-1}{x(x+h)}, h\ne 0[/tex]
