FUNCTIONS

[gkagawa]

TRUE OR FALSE? When two functions are divided, the domain of the combined function consists of all of the values in the domains of the original functions.

I'm not sure about this one, but I think this could be false, if you have the functions y=(x+3)(x-4)/(x+3), and then you get a hole (undefined value at x=-3) in the function.

 

[Deleted member 4993]

TRUE OR FALSE? When two functions are divided, the domain of the combined function consists of all of the values in the domains of the original functions.

I'm not sure about this one, but I think this could be false, if you have the functions y=(x+3)(x-4)/(x+3), and then you get a hole (undefined value at x=-3) in the function.

You have an excellent counter-example - and your deduction is correct.

 

[soroban]

Hello, gkagawa!

TRUE OR FALSE?

When two functions are divided, the domain of the combined function
consists of all of the values in the domains of the original functions.

I'm not sure about this one, but I think this could be false. . . Right!

\(\displaystyle \text{if you have the functions: }\:f(x) \:=\x+3)(x-4)\,\text{ and }\,g(x) \:=\:x+3}\)

. . \(\displaystyle \text{then }\:H(x)\:=\:\frac{(x+3)(x-4)}{x+3}\:\text{ has a hole (undefined value at }x=-3).\)

Good counter-example!

An interesting problem . . . it got me thinking (which is usually a Good Thing).
I suppose the simplest counter-example is something like this:

. . \(\displaystyle \begin{array}{cccc} \text{Function} & \text{Domain} \\ \\ f(x) \:=\:2 & \text{all real }x \\ \\ g(x)\:=\:x-1 & \text{all real }x \\ \\ H(x) \:=\:\dfrac{2}{x-1} & x \neq 1 \end{array}\)