How do people find Inverse Functions?? (1 Viewer)

[kpq_sniper017]

Easy? Average? Hard?

So far, I'm finding inverse functions are giving me a little bit of trouble. Mainly the sketching of inverse trig. functions and finding the domain of some of them.

Anyone got any tips for "inverse functions"? I could do with a few pointers.

 

[babygoose!]

well the domain of the inverse is the range of the original and the range of the inverse is the domain of the original...easiest way to remember and find the domain and range..!

 

[untamedanimal]

Its not that hard. If you know the range of a function then you know the domain of the inverse function. Getting the inverse functions is just swapping the x and y values and then making y the subject again. The hardest would be inverse trig, but thats not that hard either

 

[Collin]

They are pretty easy.

All you have to do is to limit the range of f(x) to monotonically increasing or decreasing so that there is an function of the inverse. To see if theres an inverse function simply apply the horizontal line test.

Like babygoose said, for a monotonically in or de function, the domain becomes the range and the range becomes the domain.

A function is only an inverse if it is an injective function (one-to-one uniqueness i.e mon. in or de) and a surjective function (range = codomain).

Sketching is very easy, simply mirror it around the mediator y = x.

 

[kpq_sniper017]

Just on that, how do you restrict the domain without drawing the function first? Can you test for f'(x)>0 and/or f'(x)<0??

The basic inverse trig. functions are fairly simply to sketch. It's some of the other ones that I find more difficult to get my head around:

E.g. cos^-1(1-x²/1+x²)
Domain: can anyone tell me an easy algebraic way of determining the domain?

Would you agree that practice is what makes it easier? I found it fairly confusing at first, but after practising some questions, it's becoming easier - after familiarising myself with the various questions.

 

[George W. Bush]

For your example:

We know for all x, -1<=cosx<=1.
so if y = cosx, y ranges between -1 and 1 inclusive.
now, you can see that x=arccosy, so the value of y cannot be outside that range of values.

 

[~*HSC 4 life*~]

im doin it atm, but i find it pretty hard, my teacher goes pretty fast so its like *whooosh* but eventually i hope to get there

 

[Calculon]

Tip for inverse trig: switch it around and make x the subject, it helps with domain/range.

 

[kpq_sniper017]

Originally posted by Calculon
Tip for inverse trig: switch it around and make x the subject, it helps with domain/range.

e.g.?

 

[jesshika]

hmm yeah i also have trouble with finding the restrictions for the domain on the original function in order for it to have an inverse ..
...
integration of invesrse functions
such as for eg
/
|sin^-1 x dx
/
is not ext 1 work is it not? >_<

 

[clive]

Originally posted by jesshika
integration of invesrse functions
such as for eg
/
|sin^-1 x dx
/
is not ext 1 work is it not? >_<

This is 4 unit work - Integration by parts.

Question 7 a) of 3 unit 2002 HSC has a pretty good inverse functions question to try

 

[jesshika]

i thought so ...

hey ... thanks clive ^^ ..

 

[CM_Tutor]

Originally posted by pcx_demolition017
E.g. cos^-1(1-x²/1+x²)
Domain: can anyone tell me an easy algebraic way of determining the domain?

By inspection, the domain is all real x, and the range is {y : 0 < y <= pi / 2}

How did I do it? Consider the expression we have here, x² / (1 + x²). The numerator is obviously always less than the denominator. The denominator is always positive, and the numerator is always non-negative. Thus, we have:

0 <= x² / (1 + x²) < 1, from which we can easily get:
0 < 1 - x² / (1 + x²) <= 1.

The domain and range follow.

How to do this algebraically:

We need to solve the inequality -1 <= 1 - x² / (1 + x²) < = 1, as otherwise there is no inverse cos.
-1 <= [(1 + x²) - x²] / (1 + x²) <= 1
-1 <= 1 / (1 + x²) <= 1

Now, 1 / (1 + x²) can never be negative or zero, so we actually have:

0 < 1 / (1 + x²) <= 1, which means we need
1 + x² => 1
x² => 0

Thus, x can be any real. Identifying the range follows.

 

[kpq_sniper017]

with questions like that, can you say "by inspection....." without giving any reasoning and still get full marks?

just with finding the range.....can you determine the range of a function just by observing the domain of the same function? it's easy to do with other functions, such as parabolas etc. but can you do it with inverse functions as well?

 

[CM_Tutor]

Depends on the question - If it said "State the domain and range of ..." then only an answer is required. If the question (implicitly or explicitly) required working, then (obviously) you need to show working.

 

[Calculon]

Originally posted by pcx_demolition017
e.g.?

y=3sin^-12x
sin(y/3)=2x

then you write the domain and range as if it was a regular trig function:
-pi/2< y/3 <pi/2
-3pi/2< y <3pi/2

-1< 2x <1
-1/2< x <1/2

 

[kpq_sniper017]

oh ok.....ic now.

sometimes i confuse myself though. i've been doing inverse functions for about a week and a half now. inverse trig for about a week....i still just have to get used to the domain and range of the inverse functions.

btw. in some questions u have to solve for y e.g. y=sin^-11. in these questions, if no domain or range is specified, should u use the general solutions, or should u just take the range as being -pi/2 <= y <= pi/2? (or whichever range, depending on the function)

 

[CM_Tutor]

There is no need to specify a domain or range for y = sin^-1x, as one is implied by the inverse sine function. A domain / range would need to be specified for x = sin y, but this isn't the same thing as y = sin^-1x.

 

[elizabethy]

inverse function was one of my fav topics of 3u