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well we no that the answer is f/4

there is no equal sign

your question is not complete. you need to tel us what f(x) is defined to be.

I have to explain why why f(1) = 3, f(0) = 2, and f(-1) = 1, but why f(1/4) is defined

*undefined

have you been given the expression for f(x) in terms of x?

in the 4x^2+ 7x - 2/ 4x-1

ok - now it makes sense

i simplified it to x + 2/ 1 XD

if you look at the last term, you have a division by "4x-1".
what will "4x-1" equate to when x=1/4?

\[f(x)=4x^2+7x-\frac{2}{4x-1}\]

ohhh it would be 0

correct - so you would be dividing by zero - which is undefined.

BTW: your simplification to x + 2/ 1 is not correct.

ahhhh :( what I do wrong?

I factored to get (x+2)(4x-1)/4x-1 then canceled out common factors and got x+2/1

your factorisation in wrong. remember the last term is a fraction.

\[\begin{align}
f(x)&=4x^2+7x-\frac{2}{4x-1}\\
&=\frac{(4x-1)(4x^2+7x)-2}{4x-1}
\end{align}\]

its supposed to be a rational expression o.O

it looks like this :
4x^2+ 7x - 2
--------------
4x-1

so the equation is supposed to be:\[f(x)=\frac{4x^2+7x-2}{4x-1}\]

which can be factored as:\[f(x)=\frac{(x+2)(4x-1)}{4x-1}\]

yes :)

but you cannot cancel the 4x-1 because it would be like dividing top and bottom by zero when x=1/4

i just canceled out common factors that what my teacher told me :(

you can only do that if you are told that x can NEVER be equal to 1/4

in this case I believe x can take on ay real value from minus infinity to plus infinity

and it is undefined at x=1/4

was the question asking you why is this undefined at x=1/4?

yes

so then you can state that it undefined at x=1/4 because at x=1/4 we would be dividing by zero.

okay Thank you!!! XD Wish i could giv you more than one medal o.O lol

You're welcome - I'm glad I was able to explain it well enough for you.