Question

Need help with this problem

Answer 1

Yes, what is the problem. Perhaps I know how to help...

Answer 2

Typed to get the attachment working:)

Answer 3

Thanks you!

Answer 4

well, lets plug our numbers in.
First, can you find f(1) ?

Answer 5

\(\Large\color{black}{ f(x)= \frac{1}{x+2} }\)

then, \(\Large\color{black}{ f(1)= \frac{1}{(1)~+2} =? }\)

Answer 6

what is f(1)?

Answer 7

1/3 ?

Answer 8

yes.

Answer 9

now, we are going to plug in `1+h` for x.

Answer 10

\(\LARGE\color{black}{ f(1+h)= \frac{1}{(1+h)+2} }\)

Answer 11

what does this reduce to?

Answer 12

1/3+h ?

Answer 13

yes,
\(\LARGE\color{black}{ f(1+h)= \frac{1}{h+3} }\)

Answer 14

so I should just subtract them now?

Answer 15

So, we can re-write, \(\LARGE\color{black}{ \frac{f(1+h)-f(1)}{h} }\) as, \(\LARGE\color{black}{ \frac{\frac{1}{h+3}-\frac{1}{3}}{h} }\)

Answer 16

find the common denominator, and subtract the fractions.

Answer 17

so when i subtract i get 9 + 3h as the common denominator

Answer 18

and I get h/9+3h all over h

Answer 19

So, we can re-write, \(\LARGE\color{black}{ \frac{f(1+h)-f(1)}{h} }\) as, \(\LARGE\color{black}{ \frac{\frac{1}{h+3}-\frac{1}{3}}{h} }\)

Answer 20

common denominator between \(\large \sf \frac{1}{h+3} \) and \(\large \sf \frac{1}{3}\) is \(\sf 3(h+3) = 3h +9\) you are right :)

Answer 21

and you have \(\large\sf \frac{\dfrac{h}{h+3}}{h}\)

Answer 22

thanks @Jhannybean ! is it h/h(9+3h)

Answer 23

Yes that's correct.