Question

Please help!! Let f(x) be the function 1x+12. Then the quotient
f(3+h)−f(3)/h can be simplified to −1ah+b for:
a=
and
b=

Answer 1

This is exactly what the question says?

Answer 2

yes

Answer 3

-1/ah+b sorry

Answer 4

It's just that this seems to be a difference quotient problem with a linear equation. A secant line on a linear graph is simply the original function because the secant line will pass through all points of the original line. The difference quotient is the following:

\[y=\frac{ 1(x+h)+12-(x+12) }{ h }\]

All terms cancel leaving

\[y=\frac{ h }{ h }\]

Confusing problem.

Answer 5

Ops sorry it was also
1/x+12

Answer 6

is it still working? @evansda

Answer 7

hope i get the solution :)

Answer 8

Is it (1/x)+12 or 1/(x+12) ?

Answer 9

1/(x+12)

Answer 10

does it make sence more?

Answer 11

Ok, I cannot figure this one out. There should be an x in the answer, but the problem only mentioned -1/(ah-b). Hopefully someone can see what I am missing.

Answer 12

\[f(x)=\frac{1}{x+12}\]\[\frac{f(3+h)-f(3)}{h}\]\[f(3+h)=\frac{1}{(3+h)+12}=\frac{1}{15+h}\]\[f(3)=\frac{1}{(3)+12}=\frac{1}{15}\]\[\frac{\frac{1}{15+h}-\frac{1}{15}}{h}\]\[\frac{1}{15+h}-\frac{1}{15}=\frac{15-(15+h)}{15(15+h)}=\frac{-h}{15(15+h)}\]\[\frac{\frac{-h}{15(15+h)}}{h}=\frac{-1}{15(15+h)}=\frac{-1}{15h+225}\]\[\frac{-1}{ah+b}\]\[a=15, \ b=225\]

Answer 13

Ah!!! Good work! Totally missed that.

Answer 14

Thank you kind people!!