Hello

 

could you please explain me with easy words how do you find out the domain and range of a function, for example

 

f(x) =(x+5)^3+2

 

how would you know the domain and range of that function?

 

thank you

perhaps the easiest way is to graph it. Are you allowed to do that?

Let f(x) = y then, then to get the range of y =(x+5)^3+2 we need to know what (x+5)^3 + 2 is restricted to, since (x+5)^3 can equal any real number, then so can that +2, therefore y is all real y, then to get the range:

 

y = (x+5)^3+2

 

Make x the subject

 

x = (y-2)^(1/3) - 5

 

Just looking at this, you know that (y-2) is all real numbers ®, and R^(1/3) is all real numbers and so is R - 5 is all real numbers therefore x is all real numbers.

 

Look for occasions where exponents or roots are even, that is where you will often encounter restrictions.

Also when you know some functions can't pass a certain value, eg Sin and Cos. Functions of the type you described, polynomials, are always continuous and have an infinite domain and range. When roots, or special functions are involved, then you have to check.

polynomials, are always continuous and have an infinite domain and range

 

Unless the highest power is even, in which case it will have a global minimum or maximum.

Thanks for the help everybody

 

so you are saying that if the exponent is an odd number, in this case the domain and range will be all the real numers?

 

And how can u find out the domain and range by graphing?

 

thank you!

Yes, polynomials are always defined for all values of x (hence the infinite domain), and if the highest power is odd then it will tend to infinity in one x-direction and -infinity in the other.

Graphing a function makes this clearer, for instance this particular function graphs as:

From this it is pretty clear that all y-values are defined, but alternatively the function:

f(x) = x^4 -2x^3 - 5x^2 (highest power is 4, even):

Clearly this has a minimum which will need to be investigated.

With for example [math]\frac{1}{\sqrt{x-2}}[/math], you have to be careful. You can't have a negative number in the square root, so the domain is [math][2,\infty][/math].

 

Also it will have an asymptote at y=0.

[math]lim_{x \rightarrow \infty}f(x)= 0[/math].

Actually, small mistake, x can't equal 2 either otherwise the denominator is division by zero. No negative, or zeros under a square root...

Yeah, sorry the domain should be [math](2,\infty)[/math].

 

Thanks Ragib.