Domain of multiple functions

[hdirgo]

Let f and g be two functions defined by





for any real number x.

Find and then find all the values tat are NOT in the domain of .

If there is more than one value, separate them with commas.

I got

(15-5x)
(x²-2x-24) and x=6; x=-4

 

[daon2]

I agree with your x-values for the second half. \(\displaystyle \dfrac{f}{g}(-5)\) does not mean \(\displaystyle \dfrac{f}{g}\) times \(\displaystyle -5\). It means evaluate the function (f/g)(x) at x=-5.

 

[hdirgo]

I agree with your x-values for the second half. \(\displaystyle \dfrac{f}{g}(-5)\) does not mean \(\displaystyle \dfrac{f}{g}\) times \(\displaystyle -5\). It means evaluate the function (f/g)(x) at x=-5.

So ... the solution would be -(5f)/g

 

[daon2]

So ... the solution would be -(5f)/g

No, again it is not multiplication. Do you know what "evaluate at x=-5" means?

 

[hdirgo]

No, again it is not multiplication. Do you know what "evaluate at x=-5" means?

Obviously I am not grasping the concept. :-(

 

[daon2]

\(\displaystyle \dfrac{f}{g}(x) = \dfrac{f(x)}{g(x)}=\dfrac{x-3}{(x+4)(x-6)}\)

Now plug in -5 for x.

 

[HallsofIvy]

\(\displaystyle \dfrac{f}{g}(x) = \dfrac{f(x)}{g(x)}=\dfrac{x-3}{(x+4)(x-6)}\)

Now plug in -5 for x.

Alternatively, "evaluate" f and g separately and then divide:
f(-5)= -5- 3 and g(-5)= (-5- 3)(-5+ 4)