Question

how do i find the difference quotient using
f(x) - f(a)/x-a?
With x^3 at a=3

Answer 1

your function is f(x)=x^3
and your a is 3,

so tell me 2 things so that we can proceed.

f(3)=?
f(x+h)=?

Answer 2

what you know tell me, what you don't, ask.

Answer 3

well really i cant figure much out

Answer 4

\(\large\color{slate}{ f(x)=x^3 }\), so if I wanted to find f(4), this is what i would do.

\(\large\color{slate}{ f(4)=(4)^3 }\)
\(\large\color{slate}{ f(4)=64 }\)

Answer 5

f(3) = 27

Answer 6

yes.

Answer 7

(x+h)^3

Answer 8

yes.

Answer 9

\(\large\color{slate}{ f(x+h)=(x+h)^3 }\)

Answer 10

So lets write the quotient.

Answer 11

i cant get passed (x^3)-27/(x-3)

Answer 12

the full x+h quotient?

Answer 13

yes

Answer 14

\(\large\color{slate}{ \displaystyle\frac{x^3-27}{x-3} }\)

Answer 15

factor the top, can you do that?

Answer 16

no, no x+h, that was my mistake, siorry and disregard the x+h

Answer 17

\(\large\color{slate}{ \displaystyle\frac{x^3-27}{x-3} }\) is right

Answer 18

no. that's where i'm stuck...

Answer 19

i can't figure out how it factors...

Answer 20

\(\large\color{slate}{ \displaystyle a^3-b^3=(a-b)(a^2+ab+b^2) }\)

Answer 21

now, \(\large\color{slate}{ \displaystyle\frac{x^3-3^3}{x-3} }\)

Apply \(\large\color{slate}{ \displaystyle a^3-b^3=(a-b)(a^2+ab+b^2) }\) to the top.

Answer 22

so (x-3) (x^2 +3x +9)

Answer 23

questions?

Answer 24

well thats what i kept getting but it looked wrong for some reason.

Answer 25

so the simplified answer comes out to be x^2 +3x +9?

Answer 26

yup

Answer 27

thank you very much :)

Answer 28

i hope i remember that factoring rule later

Answer 29

btw, if you are taking calculus,

\(\large\color{slate}{ \displaystyle\lim_{x \rightarrow ~a}\frac{f(x)-f(a)}{x-a} }\) is a derivative of a function f(x) at x=a, or in other words f'(a).

Answer 30

whether you are in calc or not, yw.....

Answer 31

\(\large\color{slate}{ x^2 +3x +9 }\), but i think it isn;t over

Answer 32

your x=3, right, so you have to plug that in.

Answer 33

well a=3

Answer 34

you couln't plug in 3 for x, becuase you would be getting a zero in the denominator, but now you can.

Answer 35

yes, so
\(\large\color{slate}{ x^2 +3x +9 }\) at x=3, means that
\(\large\color{slate}{ (3)^2 +3(3) +9 }\)

Answer 36

but it would just be 27
i dont think thats what she wants....

Answer 37

it is not 27

Answer 38

you are plugging 3 into the quotient, not just finding the f(3)

Answer 39

we need to evaluate \(\large\color{slate}{\displaystyle\frac{x^3-27}{x-3}}\) at x=3. We couldn't do that, because as we plug in 3 we get 0 for the denominator.


So we went ahead and factored the numerator, and got:

\(\large\color{slate}{\displaystyle\frac{(x-3)(x^2 +3x +9 )}{x-3}}\)

now x-3 cancels out. and you remain with \(\large\color{slate}{\displaystyle x^2 +3x +9 }\)

Answer 40

and \(\large\color{slate}{\displaystyle x^2 +3x +9 }\) can surely be evaluated at x=3

Answer 41

any confusion ?

Answer 42

which is 27...?

Answer 43

9 + 9 + 9 = 27

Answer 44

oh, yes, right

Answer 45

very good.

Answer 46

27 is indeed the cprrect ans.

Answer 47

don't panic if it is too simple, lol. It is a good thing not a bad thing.