ssNake's Methods 1/2 Q's.

[REBORN]

I fail to understand Mathematical Methods - requesting any help!

Q: Algebraically find the inverse function of (x+2)^2 + 3 given the domain (-infinity,-2]

I can swap x and y's and get up to the stage of y = plus/minus [ sqroot of (x-3) ] -2

But how do I know what graph to use - the plus or the minus? And how do I find the dom/range of the inverse?

[Lasercookie]

I'm pretty sure what you have there is the answer, as all the question asked was find the inverse function, not graph/sketch it.

However, if you did want to sketch it, the function you end up with is a square root function.


The simplest square root function is:

Have a go at graphing (use the CAS calculator if you can't be stuffed doing it by hand) that (if you haven't already covered this in class).

They follow the rule:

where:

• a - the dilation factor
• h - the horizontal translation
• k - the vertical translation

If a is negative, than the graph will be reflected in the x-axis.

edit: removed my domain/range stuff; got it completely wrong

[REBORN]

Hey laseredd - thanks for replying!

But I think I didn't make myself clear enough. The question says to write the rule of the inverse function with it's domain.

I essentially have two inverse functions (plus or minus) and my question is how do I know which one to choose?

My supplementary Q was how do I find the domain of the inverse?

[taiga]

Think about it logically. Imagine the initial graph with a domain from -2 to -infinity. If you flipped the x and y axis, where do you think the graph would lie?

[REBORN]

I am incapable of flipping axes in my head. Any algebra way?

[Lasercookie]

I would flip to the square root function part of your textbook and get an idea of what they look like visually if you're having trouble with it.

It should become clearer once you understand how a square root function will dilate, translate etc.

[luken93]

Domain f(x) = Range f'(x)
Range f(x) = Domain f'(x)

Find the range to the original with the domain restrictions, and this should give you the help you need to determine if it is the + or -

[sajib_mostofa]

Domain f(x) = Range f'(x)
Range f(x) = Domain f'(x)

Find the range to the original with the domain restrictions, and this should give you the help you need to determine if it is the + or -

+1

Drawing the graph should help you.

[xZero]

I am incapable of flipping axes in my head. Any algebra way?

think of it this way, your y-axis becomes the x-axis, this logically implies that the restriction from y-axis carries over to the new x-axis. Same thing with the old x-axis.

[REBORN]

Hmm okay.

Q: y = 3 x [ sqroot (8-2x) ] +5

How do we find the maximal domain/range WITHOUT sketching the graph?

For domain we let 8-2x = 0 ---> 8 = 2x ---> x = 4

For range we --->   

[b^3]

so domain is -infinty to 4 so sub values in for range, so its a postive graph but flipped in y-axis so min value will be when x=4 so y=3srt(0)+5=5 and for the max it will keep on going to infity so range is [5,infinity)

[pi]

Hmm okay.

Q: y = 3 x [ sqroot (8-2x) ] +5

How do we find the maximal domain/range WITHOUT sketching the graph?

For domain we let 8-2x = 0 ---> 8 = 2x ---> x = 4

For range we --->   

I think that for the domain, try letting instead... Sub the endpoint for the range.

EDIT: supposed to be 'greater or equals to' sign

[brightsky]

domain:



range:

[REBORN]

Have no clue what you did for range there?

You start with the inside of the sq root brackets and just added each transformation...?

:S

[brightsky]

we have the equation
now look for a restriction in the equation that we are absolutely sure of.

we know that . always for the appropriate x.

hence now we work with this inequation:



multiply 3 to both sides:



add 5 to both sides:



note that i'm not adding transformations, i'm merely exploiting basic arithmetic rules. (say for instance you have x>0, you can always say that x+1>1, etc.)

now coincidently

so sub that in:



hope this clears things up a little