Prelim 2016 Maths Help Thread (1 Viewer)

[eyeseeyou]

There's that attitude again. Don't get people to help you only to then use harsh language and curse words later

The graph of y=f(x) i.e. y=tan(x/2) is simply one branch of the tan curve with period 2pi.

On this domain, since f(x) is one-to-one it is indeed invertible. To find the inverse:

Consider x=tan(y/2)
Then arctan(x)=y/2 so y=2 arctan(x)

where arctan is just inverse tan if you have not seen it before. I.e. tan^-1(x)

So we have f^-1(x) = 2arctan(x)

This is a dilated version of the regular inverse tangent curve: the asymptotes are now at pi and -pi, instead of pi/2 and -pi/2
View attachment 33257

Leehuan as I said above could you please help me with this question (i.e. solve it wihtout arctan since idk what that is)

 

[eyeseeyou]

ANother inverse function question I am struggling with

1. Find the inverse function of the following. Also state the domain and range of the inverse function
a. y= (x+1)^2 -1, x is greater than or equal to 2
b. y=(x-2)^2 -1, x is greater than or equal to 2
c. y=8(x-1)^3

So for a and b I found out the equation and found the domain but am struggling to find the range

and for c I just can't seem to solve it for some reason

Um....This is sorta urgent, could you please help me

 

[leehuan]

Leehuan as I said above could you please help me with this question (i.e. solve it wihtout arctan since idk what that is)

I'm confused to what you mean. Arctan is just another way of writing inverse tangent.

If you don't know what the tan^-1 button is on your calculator then I have no idea how to help you.

The graph was shown anyway. It is obvious that the graph of an inverse function is just the graph of the original function, rotated about the line y=x. This is something you should've already known before attempting inverse trig in the whole inverse functions topic.

 

[leehuan]

Um....This is sorta urgent, could you please help me

There is nothing hard about here. Consider a)

 

[eyeseeyou]

I'm confused to what you mean. Arctan is just another way of writing inverse tangent.

If you don't know what the tan^-1 button is on your calculator then I have no idea how to help you.

The graph was shown anyway. It is obvious that the graph of an inverse function is just the graph of the original function, rotated about the line y=x. This is something you should've already known before attempting inverse trig in the whole inverse functions topic.

Oh yeah dw, just don't say arctan next time b/c I'm not used to these terms

 

[eyeseeyou]

I tried this: y=2-2e^-2x

I got up to 2e^-2y=5-x

I am so confused??? what do I do

 

[eyeseeyou]

There is nothing hard about here. Consider a)

How does the range get resitricted to 8? Is this wen you sub x is greater than or equal to 2 in the eqn?

 

[leehuan]

Oh yeah dw, just don't say arctan next time b/c I'm not used to these terms

You're gonna see that a LOT if you do more maths in the future.

How does the range get resitricted to 8? Is this wen you sub x is greater than or equal to 2 in the eqn?

When in doubt draw the graph.

Then I learn how to just picture it mentally.

 

[leehuan]

I tried this: y=2-2e^-2x

I got up to 2e^-2y=5-x

I am so confused??? what do I do

Are you also trying to invert this one?

 

[eyeseeyou]

what is the domin of -x+ square root of (x+3)?

I got x is greater than or equal to -3 (Domain) and y is greater than or equal to 0 (range)

 

[eyeseeyou]

for this eqn y=5+ squareroot of (3-x) I got y=28-x^2 and apparently that's wrong

 

[leehuan]

what is the domin of -x+ square root of (x+3)?

I got x is greater than or equal to -3 (Domain) and y is greater than or equal to 0 (range)

Upon closer inspection, this one is a bit harder.

 

[InteGrand]

for this eqn y=5+ squareroot of (3-x) I got y=28-x^2 and apparently that's wrong

You probably just made a silly mistake in expanding something. Also make sure to take note of the range and domain of the inverse.

 

[eyeseeyou]

You probably just made a silly mistake in expanding something. Also make sure to take note of the range and domain of the inverse.

Please help me

 

[leehuan]

Please help me

If you need a correction, can you please at least include your progress thus far so we can tell you WHERE you went wrong

 

[InteGrand]

Please help me

How can I help you?

 

[eyeseeyou]

for this eg f(x)=-x^2, x is greater than or equal to 0, why does the mirroring of the graph occur at the origin (graph it so you see what I am talking about)

BTW Integrand, my physics questions are still unanswered

 

[eyeseeyou]

Also am doing this wrong for some reason

f(x)=x^2+6x+5 where x is less than or equal to -3

 

[InteGrand]

for this eg f(x)=-x^2, x is greater than or equal to 0, why does the mirroring of the graph occur at the origin (graph it so you see what I am talking about)

BTW Integrand, my physics questions are still unanswered

What do you mean mirroring of the graph? Graph of what? The inverse's graph is the reflection about the line y = x of the original graph.

 

[InteGrand]

Also am doing this wrong for some reason

f(x)=x^2+6x+5

What do you think that reason is?