Fur Maths Theory of Functions

[Hrishi98]

Could anyone in Math HL/SL explain to me in brief, how to identify domain and range of a particular equation.

Any good method to easily identify them 

Example:-

 

f(x)= 1/x+2/x+1

[ctrls]

You've tagged this post to be about further maths, but I'm assuming this is from HL/SL maths. What I'm about to say strictly speaking isn't correct (as you would see if you do the sets, relations and groups chapter), but this holds for the treatment of functions in the respective core syllabi.

 

The functions you have to consider are generally a composition of basic functions, which may be added and/or multiplied together, etc. Normally it is assumed that the function can take any real number as an input (so the domain is all the real numbers), but as you know some of these functions may not be defined for specific values, of x, notably that:

• 1/x is undefined at x=0
• log x is undefined for x=0 and x<0
• sqrt(x) is undefined for x<0
• tan is undefined for x = Â±pi/2, Â±3pi/2, ...
• sin^-1 and cos^-1 are undefined for x>1, x<-1

Off the top of my head, I believe these are the only cases that you could be reasonably asked about in the IB. When it occurs that the value of f(x) is undefined at a particular value of x, we say the domain doesn't contain this particular x. Hence the function you stated, as notated above, has a domain of all the real numbers except for 0.

 

It is worth noting that it's a bit more complicated than this, a more proper treatment requires the consideration of limits and a proper definition of what a function is, but for our purposes this is sufficient.

 

Generally sketching the curve is the clearest way of finding the range, identifying the min/max points and asymptotes is usually enough to get an idea of what the curve looks like and can then see what values of y are not touched. Of course, it isn't necessary to sketch a curve if it's clear what it is by some thinking, like schulzeug demonstrated above. Knowing the ranges of the various elementary functions (trig functions, log, roots, etc) is helpful here, as is knowledge of transformations of functions and what they do to the range (ie: how f(x)+c, c f(x), f(x+c) respectively change the range).

Edited December 28, 2014 by ctrls

[Hrishi98]

Its actually 

f(x)= (1/x)+(2/x+1)

 

And so overall from the explanation you provided, I must first simplify these solutions and then use the above rules you provided to find domain and then range right?

well, heres one which completely shut my brain down last night....

 

f(x)= e^x+((2x+3)^1/2)-(1/x^2+4)-2

How do I simplify this?

[ctrls]

I'm guessing you mean,

 

Simplification doesn't really work here, since you really can't simplify this much further. In general it's a first step to clear up an expression, but that alone won't really give any answers. Past that there's no real algorithmic way to solve these problems, other than to determine the nature of the function and work from there.

 

To find the domain, consider each term separately. The function e^x is well defined for all values of x, so that's fine. The fractional term will be undefined if the denominator equals 0, but this doesn't occur for any real value of x in this case. Subtracting two doesn't affect the domain either.

 

If the argument of the square root is negative however, this will be undefined if we are considering only real numbers - which is usually the case in the IB, unless otherwise stated. Hence if 2x+3<0, then the function won't be defined. So the domain of the function will be x≥-3/2.

 

Finding the range is a bit harder. I may be missing something, but I can't think of a simple way to determine the nature of the curve. The one way which comes to mind is the method I described previously, which would work as follows,

 

You can prove that the function has no stationary points, in other words show there are no solutions to the equation f'(x)=0 in the domain x≥-3/2. If this is true, you can then consider the points f(-3/2) and f(x) as x tends to infinity, which gives the range (the latter diverges to infinity, due to the e^x term, so the range is f(-3/2)≥0. If this isn't clear, try sketching a curve with no stationary points starting at x=-3/2, considering the asymptotic behaviour as x tends to infinity.

 

Problem is that showing this function has no stationary points isn't particularly easy, you get a pretty ugly and complication equation which requires a fair bit of work to show has no roots. Other than that however, I'm out of ideas.