Question

Calculus Help http://i.imgur.com/FBS2gqh.png I'm on number 52, and the limit is 1/3, can anyone help me?

Answer 1

a^2-b^2 = (a-b)(a+b)

Answer 2

The denominator is equivalent to:
\[(\sqrt{x})^2-3^2\]

Answer 3

can you elaborate more onj how to solve it? I just started derivatives today

Answer 4

You can rewrite the denominator as:
\[(\sqrt x-3)(\sqrt x + 3)\]

Answer 5

And on top, if you pull out a 2, you get:
\[2(\sqrt x-3)\]

Answer 6

this is a weird question with an infinite amount of answers for f and c

Answer 7

no

Answer 8

umm, I'm not following you bahrom, how does that relate to the problem? I'm not being snarky, I'm just not familiar yet

Answer 9

im busy atm but simplify the expression and plug in x = 9

Answer 10

u get
\[\frac{2}{\sqrt{x}+3}\]

Answer 11

and when u plug in x=9, u get:
2/(3+3)=2/6=1/3

Answer 12

I knew the limit was 1/3, I was asking about the other parts

Answer 13

I told you how to do it.. factor the denominator and numerator.. then simplify and plug in x = 9

Answer 14

he wants to find f and c given f'(c) = 1/3

there are infinitely many solutions

Answer 15

ohhhh sorry lol

Answer 16

use the limit defn of a derivative no?

Answer 17

f(x+h)-f(x)
---------
x-h

Answer 18

I thought it was f(x+h)-f(x) / h?

Answer 19

oh ure right

Answer 20

Actually i think it's the same, the diff is in your case h goes to 0, and in mine, x goes to h

Answer 21

But im not sure... honestly i have no idea how to do this, blanking out for some reason.. @Zarkon

Answer 22

\[f(x)=2\sqrt{x}\]

Answer 23

\[f'(x)=\frac{1}{\sqrt{x}}\]

\[f'(9)=\frac{1}{\sqrt{9}}=\frac{1}{3}\]