• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Algebraic Proof :) (1 Viewer)

Energised

New Member
Joined
Apr 29, 2015
Messages
7
Gender
Undisclosed
HSC
2016
Hello,

Can someone please help me with this question:

If , show that .

Thank you very much.

Energised
 

Shadowdude

Cult of Personality
Joined
Sep 19, 2009
Messages
12,145
Gender
Male
HSC
2010
I thought:


And then. . . Can you simplify this side to = the other or?
Uhh you made an algebra mistake when substituting in... because that simplification isn't right

Try on paper
 

Energised

New Member
Joined
Apr 29, 2015
Messages
7
Gender
Undisclosed
HSC
2016
Sorry changed now. . .

So did you multiply each term by x to get to the next step??

Thanks. :)
 

Shadowdude

Cult of Personality
Joined
Sep 19, 2009
Messages
12,145
Gender
Male
HSC
2010
Sorry changed now. . .

So did you multiply each term by x to get to the next step??

Thanks. :)
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
 

Energised

New Member
Joined
Apr 29, 2015
Messages
7
Gender
Undisclosed
HSC
2016
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
Sorry, getting a bit mixed up trying to use latex!!

I can get how you got to the next step... Now...
 

Energised

New Member
Joined
Apr 29, 2015
Messages
7
Gender
Undisclosed
HSC
2016
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. :)
 

Shadowdude

Cult of Personality
Joined
Sep 19, 2009
Messages
12,145
Gender
Male
HSC
2010
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. :)
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.
 

Energised

New Member
Joined
Apr 29, 2015
Messages
7
Gender
Undisclosed
HSC
2016
You have . But so the point of discontinuity occurs at .
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.

Thanks. :)
 

Energised

New Member
Joined
Apr 29, 2015
Messages
7
Gender
Undisclosed
HSC
2016
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.
Thanks I understand, (you just simplify again, then multiply each term by x.

How's the x can't = 0 on the end impact?
 

panda15

Alligator Navigator
Joined
Feb 22, 2012
Messages
675
Gender
Male
HSC
2013
Thanks I understand, (you just simplify again, then multiply each term by x.

How's the x can't = 0 on the end impact?
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.
 

Energised

New Member
Joined
Apr 29, 2015
Messages
7
Gender
Undisclosed
HSC
2016
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.
Thank you for your explanation, yes, that makes sense now. :)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top