for (i), we want all the x's in the domain of p(x) for which $\displaystyle p(x) \ne 0$
for (ii) we want all the x's in the domain of p(x) AND q(x) (at the same time) for which $\displaystyle q(x) \ne 0$
for (iii) we want all the x's in the range of q(x) that are in the domain of r(x)
for (iv) we want all the x's in the range of p(x) that are in the domain of s(x) and of course, we cannot have s(p(x)) = 0
sorry i didnt understand a single thing you just said, would you mind explaining or elaborating a bit more please?? except for the part that the denominator cant be zero..i understand that as the function will be undefined.
thanks, and also for the question i posted earlier about continuity..i got k=29...so thats the value where it would be continuous..am i correct? please check
thanks heaps
so for (ii), we need to make sure that the x's we choose work in both p(x) and q(x) at the same time, that's all i'm saying. yes, we cannot choose x's that make q(x) zero, but we can't choose x's that are in the domain of q(x) but not in the domain of p(x) either, since the top function won't work and we can't get a value for the fraction
for (iii). the range is the output values we can have. so for, say, q(x) the domain is the x's we can plug into q, while the output values q(x) are the range. for example. for the function y = x^2, the domain is all real x, because we can plug in anything we choose for x. however, the output values are "limited", no matter what we plug in, we will always get an output value greater than or equal to zero, so the range of y = x^2 is all y's such that $\displaystyle y \ge 0$ or in other words, $\displaystyle y \in [0, \infty)$. a similar thing holds here.
when we say r(q(x)), it means we are taking the outputs from the function q and plugging them into the function r. so first, we need to know what are the possible output values for q, and then we need to know, which of these values are in the domain of r. those are the values that are in the domain of r(q(x))
a similar thing holds true for (iv)
got it?
please make this comment in the thread that you are refering to, so other users can followthanks, and also for the question i posted earlier about continuity..i got k=29...so thats the value where it would be continuous..am i correct? please check
$\displaystyle (-\infty, 2/3) \cup (2/3, \infty)$
incorrect. both p(x) and q(x) are polynomials, all x's work in them. thus our only concern here is that q(x) is not zero. what x's make that happen?(ii) (-infinity, 2/3) U (2, infinity)
correct(iii) (-infinity, -2) U (2, infinity)
here we have to choose those x's that makes $\displaystyle p(x)>1$ (why is that?)(iv)..no idea
so what's the answer?
correctas i said, any x can work in both p(x) and q(x), so all we care about is that q(x) is non-zero. so we wantalso could you please give me the correct answer for p(x)/q(x)..im really tryin on that one but getting no where...
$\displaystyle q(x) = x^2 - 4 \ne 0$
$\displaystyle \Rightarrow (x + 2)(x - 2) \ne 0$
$\displaystyle \Rightarrow x \ne 2 \mbox{ and } x \ne -2$
so the domain is: $\displaystyle (-\infty, -2) \cup (-2,2) \cup (2, \infty)$