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Consider the following definitions of operations in functions: 

                       (f + g)(x) = f(x) + g(x)
                       (f - g)(x) = f(x) - g(x)
                       (fg)(x) = f(x)g(x) and 
                       (f/g)(x) = f(x)/g(x)

consider, f(x) = $\sqrt{x}$ and g(x) = $\sqrt{1-x}$,
here, the domain of f(x) = [0, ∞), and the range is also the same,
for g(x) the domain is [- ∞, 1] and the range is [0, ∞).

Now, consider the following operations from this function:

(f+g)(x)--------------> $\sqrt{x}$ + $\sqrt{1 -x}$(domain: [0,1]

(f-g)(x)--------------> $\sqrt{x}$ - $\sqrt{1- x}$(domain: [0,1]

(f.g)(x)--------------> $\sqrt{x}$.$\sqrt{1-x}$---->$\sqrt{x(1-x)}$(domain:[0,1]

(f/g)(x)------------>$\sqrt{x/(1-x)}$(domain: [0, 1)

(g/f)(x)------------->$\sqrt{(1-x)/x}$(domain: (0,1]

It is quite obvious for the domain to be the common interval of both the functions, since the operated function(e.g. (h: (f+g)(x)), must satisfy for both the intervals for both the functions.

But, What follows for the range? Below I have presented a graph of the two operations plotted on geogebra:
enter image description hereclick to enlarge, the equations are on left

Example for the operation (f-g)(x), the range is not [0,∞), nor something related to the union or intersection of the two ranges.

A bit down the list, the more interesting of all is the division, (f/g) omits the right interval of the domain and (g/f) omits the left. Is this a coincidence or a general rule, if so, is there a proof for this?

And, lastly, is there a general rule to guide the values of domain and ranges for such algebraic operations?

mathmaniage
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1 Answers1

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Consider a function h(z), such that $$h(z)=f(x)/z$$

The domain of z in this case will be:

$$z \in (-\infty,0),(0,\infty) $$ and its range will be the same

When we substitute a function f or g for z, the range of what we substitute becomes the domain of h. Making the composite function h(g(x)).

But because the domain of h(z) is also the conjunction of f and z's domain (as you observed earlier), our final functions domain will be the conjunction of f(x)'s domain, h(z)'s original domain and the range of what we substituted.

As a general rule, whenever you divide two functions; take the denominators range and form a conjunction of it with the numerators domain, and form a conjunction of that with a composite functions domain, that has the form of your new function.

Symbolically $$dom \ h(g(x))=(range \ g) \cap (dom\ h(z)) \cap (dom\ f)$$

In answer to your other remark:

A bit down the list, the more interesting of all is the division, (f/g) omits the right interval of the domain and (g/f) omits the left. Is this a coincidence or a general rule, if so, is there a proof for this?

no that symmetry is just a coincidence.

Excellent question.

user400188
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    Also, your graph of h(x) seems to be wrong; as when x is zero, h(0) has a value of -1. Although I am partially color blind so it may just be I'm interpreting the image wrong. – user400188 Jan 01 '17 at 10:28
  • no, it might be a bug in geogebra, I checked division by zero graphs later, and it plots it somehow,. Oh and thanks for your answer! – mathmaniage Jan 01 '17 at 14:54