Question

if f(x) = x^2 + x - 1, find (f(x+3) - f(3))/x, x cannot equal 0

Answer 1

would i find f(x+3) and (f(3) and plug it in?

Answer 2

\[f(3)=3^2+3-1=11\]
\[f(x+3)=(x+3)^2+(x+3)-1=9+6x+x+3-1=7x+11\]

Answer 3

yes just plug it in

Answer 4

then subtract, to get
\[f(x+3)-f(2)=7x+11-11=7x\] then divide by \(x\)

Answer 5

typo there, i meant
\[f(x+3)-f(3)=7x+11-11=7x\]

Answer 6

when i factored f(x+3) I got x^2 + 6x + 9 + x + 3 + 1?

Answer 7

don't forget so square \((x+3)^2\) correctly
yes, that is right, except in your question you have -1 not +1 i think

Answer 8

so it should be

\[x^2 + 6x + 9 + x + 3 -1\] right?

Answer 9

it is minus one, but yes. and then shouldn't the 3 simply be 3, since there is no x after the f. it's only f(3)

Answer 10

other than that it is correct. combine like terms, subtract 11

Answer 11

\(f(3)=11\) it is the number you get when you replace \(x\) by 3

Answer 12

wait nevermind you're right

Answer 13

I got the answer x + 7...

Answer 14

if you do more than a couple of these you will notice that in the numerator you are left only with \(x's\) and all the numbers add up to zero

Answer 15

ok that was wrong! i mean i was wrong, not you

Answer 16

you are correct. you get
\[\frac{x^2 + 6x + 9 + x + 3 -1-11}{x}\]
\[\frac{x^2+7x}{x}\]
\[x+7\]

Answer 17

you have it correct, i made a mistake

Answer 18

how is that the answer though?

Answer 19

then one you got? your answer is right

Answer 20

oh alright! thank you!

Answer 21

\[f(x+3)=x^2+7x+11\]
\[f(3)=11\]
\[f(x+3)-f(3)=x^2+7x+11-11=x^2+7x\]
\[\frac{f(x+3)-f(3)}{x}=\frac{x^2+7x}{x}=\frac{x(x+7)}{x}=x+7\]