1
$\begingroup$

I have searched everywhere on ways to do these Combining Functions problem so here goes. Everywhere I look, this is only explained through actual functions. Ive yet to see one done with only the domain and range of a function.

Let f,g,h be functions with domain and ranges below: f has domain [-1,1) and range [0,2) g has domain [0,2) and range [-1,1) h has domain [1,3) and range [1,2)

For each of the following proposed new functions, specify its domain if it exists, otherwise state that the function does not exist.

(f+g), (f+h), (g o h), (h o g)

Any advice would be a godsend. I already have the answers, just looking for explanations for them.

$\endgroup$
3
  • $\begingroup$ Well, take $f+g$, say. What might that mean? Well, for some $x$ we'd hope to write $(f+g )(x)=f(x)+g(x)$ but for that to make sense we'd need to have both $f(x)$ and $g(x)$ defined. Is that possible? Yes! the intersections in the domains of $f,g$ is $[0,1)$. So $f+g$ has domain $[0,1)$. I don't believe it is possible to specify the range without further information. We know it is contained in $[-1,3)$ but we don't know exactly what it is. the others are similar. $\endgroup$ – lulu Mar 15 '17 at 11:32
  • $\begingroup$ Should say: the definition of "range" isn't universally agreed on. I believe most people say it means the set of values taken by function. Others say it is simply a set that the function is said to map to, with no assumption that it hits every value. Do you know which definition you are using? $\endgroup$ – lulu Mar 15 '17 at 11:34
  • $\begingroup$ The Wiki article on Range gives a useful discussion of the ambiguity between the two possible definitions. $\endgroup$ – lulu Mar 15 '17 at 11:42
0
$\begingroup$

(f + g) has domain [0,1) and range [-1,3).

(f + h) has domain [-1,3) and range [1,4).

(g o h) has domain [1,3) and range [-1,1).

(h o g) isn't well defined.(It may or may not exist.)

Note that domain of a function is the set of all values that the function can accept. Asking for the value of a function outside its domain is absurd. The range is the super-set of the outputs for the function.

In the first two cases, for the sum of functions to be defined, I have taken the intersection of the domains of the two functions (for points outside this intersection, at-least one function is going to be not defined). I could have set the domain of the function to $\mathbb R$, but I chose the set such that it is impossible for the function to attain values outside this set.

In the third and fourth problems, for (f o g) to be well defined on the domain of g, the range of g must be a subset of the domain of f. We can also restrict the domain to only include points such that g(x) $\in$ domain(f).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.