Hello,
Can someone please help me with this question:
If , show that .
Thank you very much.
Energised
Uhh you made an algebra mistake when substituting in... because that simplification isn't rightI thought:
And then. . . Can you simplify this side to = the other or?
Yeah, I multiplied by \frac{x}{x}Sorry changed now. . .
So did you multiply each term by x to get to the next step??
Thanks.
Sorry, getting a bit mixed up trying to use latex!!Yeah, I multiplied by \frac{x}{x}
Also your edit's numerator is still wrong... I'd suggest you go from my working and figure out the intermediate steps to get what I have exactly
You mean the one starting with "But"?How do you get to the last line??
Also how about this q: (I just can't get the last part)
The graph of is the line y= x +14 (I think) with a discontinuity at the point _________?.
Thanks.
Ah, thanks heaps. That makes sense. And the reason that x can't = 1, is because the bottom of the fraction cannot = 0, am I right.You have . But so the point of discontinuity occurs at .
Thanks I understand, (you just simplify again, then multiply each term by x.You mean the one starting with "But"?
It's a separate thing. You simplify the thing on the previous line, and then you should see the link.
Because g(x) is defined at x=0, but g(1/x) is not because 1/x is not defined at x=0, so g(x) does not equal g(1/x) at x=0.Thanks I understand, (you just simplify again, then multiply each term by x.
How's the x can't = 0 on the end impact?
Thank you for your explanation, yes, that makes sense now.Because g(x) is defined at x=0, but g(1/x) is not because 1/x is not defined at x=0, so g(x) does not equal g(1/x) at x=0.