Question

Find the difference quotient and simplify:
f(x)=x^2-x+1, (f(2+h)-f(2))/(h), h cannot equal 0

Answer 1

\[f(x)=x ^{2}-x+1, \frac{ f(2+h)-f(2) }{ h }, h \neq0\]

Answer 2

slowly
what is \(f(2)\)?

Answer 3

is it \[f(x)\]

Answer 4

\[f(x)=x^2-x+1\] as you wrote
so \[f(2)=2^2-2+1=?\]

Answer 5

3

Answer 6

right good
so next question, what is \(f(2+h)\)?
hint, your answer will have an \(h\) in it
hint hint: since \[f(\clubsuit)=\clubsuit^2-\clubsuit+1\] you have
\[f(2+h)=(2+h)^2-(2+h)+1\]
needs algebra to clean it up

Answer 7

\[4+h^2-2-h+1=3+h^2-h\]
Or would the h^2 and h turn into only one h?

Answer 8

close

Answer 9

\((2+h)\neq 4+h^2\) you got to multiply out \[(2+h)^2=(2+h)(2+h)=?\]

Answer 10

4+4h+h^2?

Answer 11

yes

Answer 12

so now we are at \[4+4h+h^2-2-h+1-3\] for the top of the fraction \[f(2+h)-f(2)\]
now just a tiny bit of arithemtetic

Answer 13

notice that the numbers add up to zero, all your are left with is an expression in \(H\)

Answer 14

\(h\)

Answer 15

oh and \(h^2\)

Answer 16

So we're left with h^2+3h?

Answer 17

yes

Answer 18

that is the numerator
so we are at \[\frac{3h+h^2}{h}\] as a fraction
now divide each term by \(h\)

Answer 19

So it's 3+h!

Answer 20

yes it is

Answer 21

this a calc class or not yet?

Answer 22

Thank you so much, I struggled a lot with that problem. And it's for my pre-cal class

Answer 23

when you take calc by week 2 you till be able to come up the the answer instantly in your head

Answer 24

yw

Answer 25

I hope so