Silly Questions Thread

[the.watchman]

Yeah, for your equation notice that every x is always positive, so all you have to do is sketch for and reflect in the y axis..

Yep, but for a more general case, stuff like has an axis of symmetry not at x=0, but at x=2, so be mindful that the symmetry line is when the inside of the modulus is equal to zero.

Not that that equation will pop up in methods. Ever.

ok so would it split up like so?

Yes, you can do that, but once you get more experienced, you only need the positive side, then reflect it

[m@tty]

Yeah, I was only saying for this specific case. And in your equation there would also be zeros at

[the.watchman]

Yeah, I was only saying for this specific case. And in your equation there would also be zeros at

Lol, misconfusion galore
And No. No. No.

[TrueTears]

Yeah, I was only saying for this specific case. And in your equation there would also be zeros at

2^0 = 1

[m@tty]

I see, I thought that it was . But it is true for the mod portion, which is what I was referring to in any case.

[Aqualim]

The function with rule , will have an inverse function if;


It's a multiple choice question for the record

EDIT: made a mistake

[the.watchman]

You have to think which of those domains given by the answers makes f(x) one-to-one

I'd say it's both (a) and (d) because they are the same
For a>0, only one arm of the graph remains, and it is one-to-one

[Aqualim]

ok so would you just simply draw the graph in the calculator to know what the graph looks like or would you sketch it somehow?

[superflya]

ok so would you just simply draw the graph in the calculator to know what the graph looks like or would you sketch it somehow?

easier to graph it on calc, but u could sketch. just visualize and ull be able to see if its one to one.

[the.watchman]

ok so would you just simply draw the graph in the calculator to know what the graph looks like or would you sketch it somehow?

You can imagine it in your head, it would look similar to ln(|x|) but steeper
You should also note that there are two arms, and each arm is one-to-one, so you need to restrict the graph to just one arm.
This is allowed by a>0

[Aqualim]

I'm assuming you are able to visualise this in your head, just by being exposed to this type of graph over a longer period of time yeah? cause I can't just visualise how a 'g of f(x)' would look off the top of my head

[the.watchman]

I'm assuming you are able to visualise this in your head, just by being exposed to this type of graph over a longer period of time yeah? cause I can't just visualise how a 'g of f(x)' would look off the top of my head

Lol, just think of it as x^4 is 'similar' to x but it increases heckloads quicker and starts off slower, right?
And the negative side is the positive side reflected

[Aqualim]

I'm assuming you are able to visualise this in your head, just by being exposed to this type of graph over a longer period of time yeah? cause I can't just visualise how a 'g of f(x)' would look off the top of my head

Lol, just think of it as x^4 is 'similar' to x but it increases heckloads quicker and starts off slower, right?
And the negative side is the positive side reflected

Makes sense, also just found out by plugging various powers into the graphs part of the calculator that for every odd power there is one arm and for every even power there are two arms, just as the power increases the graph is stretched more in the y-direction

[Aqualim]

For
show that , where u and v are real numbers

[the.watchman]

AND