Why would f(1/4) be undefined?
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?
can you please try and explain to me step by step what you did so that I can spot where you went 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.
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