Question

find the difference quotient for 6/(x^2)

Answer 1

???

Answer 2

he wants to find \[\large \frac{f(x+h)-f(x)}{h}\] where in this case \[\large f(x)=\frac{6}{x^2}\]

Answer 3

yeah i dont know what to do once i have \[(6/(x^2+2xh+h^2) -6/x^2)/h\]

Answer 4

multiply EVERY term by the inner LCD \[\large x^2(x^2+2xh+h^2)\] this will clear out the inner fractions.

Answer 5

so then you get (6x^2(x^2+2xh+h^2)-6x^2(x^2+2xh+h^2))/h?

Answer 6

not quite, it will give you


\[\large \frac{6x^2-6(x^2+2xh+h^2)}{hx^2(x^2+2xh+h^2)}\]

Answer 7

what's next then?

Answer 8

0/h?

Answer 9

distribute the 6 through to get \[\large \frac{6x^2-6x^2-12xh-6h^2}{hx^2(x^2+2xh+h^2)}\]


notice the 6x^2 terms cancel, so we then get \[\large \frac{-12xh-6h^2}{hx^2(x^2+2xh+h^2)}\]


now factor out an h from the numerator \[\large \frac{h(-12x-6h)}{hx^2(x^2+2xh+h^2)}\]


this h up top cancels with the h down below to give us \[\large \frac{-12x-6h}{x^2(x^2+2xh+h^2)}\]


From there, we can distribute the terms in the bottom to give us \[\large \frac{-12x-6h}{x^4+2x^3h+x^2h^2}\]


and this is as far as we go since we can't do anything with h (not yet at least)

Answer 10

i dont know how to move on

Answer 11

that's the last step

Answer 12

i dont know how to simplify though cant it be simplified?

Answer 13

if you're familiar with limits, you would then plug in h=0 and simplify to get \[\large -\frac{12}{x^3}\]

Answer 14

but since I'm assuming you're not, the last step is then \[\large \frac{-12x-6h}{x^4+2x^3h+x^2h^2}\]

Answer 15

alright thanks