What am I supposed to do here?

[Loki123]

solution 1 = -1
solution 2 = 13/2
final solution [-1,13/2]

this is not correct, so i guess i do not understand.

 

[blamocur]

this is not correct, so i guess i do not understand.

In your post #8 you correctly figured out two intervals, i.e., [imath][0,1][/imath] and [imath][-4,0][/imath], one for each of the two parts of the new functions. As @Steven G pointed out, your new domain contains [imath]x[/imath]s which belong to both intervals so that you can compute both parts.

 

[Loki123]

In your post #8 you correctly figured out two intervals, i.e., [imath][0,1][/imath] and [imath][-4,0][/imath], one for each of the two parts of the new functions. As @Steven G pointed out, your new domain contains [imath]x[/imath]s which belong to both intervals so that you can compute both parts.

that was incorrect #15 is correct

 

[blamocur]

that was incorrect #15 is correct

I disagree: both [imath]2x[/imath] and [imath]\frac{x}{2}+2[/imath] must be in [imath][0,2][/imath]

 

[topsquark]

solution 1 = -1
solution 2 = 13/2
final solution [-1,13/2]

We've all but told you the answer to this. There is a simple fact you aren't remembering.

The domain of f is [0, 2].

So if you are calculating that the domain using (x/2) + 2 then (0/2) + 2 = 2 and (2/2) + 2 = 3 and 2 < 3. What values of x are allowed?

That's as far as I'm prepared to give you hints.

-Dan

 

[Loki123]

We've all but told you the answer to this. There is a simple fact you aren't remembering.

The domain of f is [0, 2].

So if you are calculating that the domain using (x/2) + 2 then (0/2) + 2 = 2 and (2/2) + 2 = 3 and 2 < 3. What values of x are allowed?

That's as far as I'm prepared to give you hints.

-Dan

hm
so here is the thing
i am inclined to say only 2 because of the domain of f you mentioned, however, i am unsure as to why that would be...
we calculated the new domain for new functions.

 

[Steven G]

You can compute f of any number between 0 and 2 inclusive. That was given.
Now you can replace x in f(x) with any argument you like but that argument still has the constraint 0 to 2.
If you call the argument 2x, ie if you have f(2x), you can not use x=1.5. Why not? Because 2x would be 3 and you can’t compute f(3) since f(x) is not defunded when the argument is outside of 0, 3.
That is all the help that I can give you.

 

[topsquark]

hm
so here is the thing
i am inclined to say only 2 because of the domain of f you mentioned, however, i am unsure as to why that would be...
we calculated the new domain for new functions.

Let's try this. For what values of x is (x/2) + 2 in [0, 2]?

-Dan

 

[Loki123]

I disagree: both [imath]2x[/imath] and [imath]\frac{x}{2}+2[/imath] must be in [imath][0,2][/imath]

I think you were correct since now I have a correct answer.

 

[Steven G]

I'll do the 1st one for you.
0<x<2. Now double all three terms. This gives you 0<2x<4

I am sorry but what I wrote above is complete nonsense. Sorry about that.

I will post the solution since I confused you.

The argument, whatever it is, must be between 0 and 2.
So 0<2x<2 => 0<x<1.
0<x/2 + 2< 2 => -2 < x/2 < 0 => -4 <x <0.

So to compute f(2x) and f(x/2 +2) we need 0<x<1 AND -4<x<0. That means, as you said, x=0

 

[Steven G]

that was incorrect #15 is correct

Actually, it was incorrect. Sorry.

 

[Loki123]

I am sorry but what I wrote above is complete nonsense. Sorry about that.

I will post the solution since I confused you.

The argument, whatever it is, must be between 0 and 2.
So 0<2x<2 => 0<x<1.
0<x/2 + 2< 2 => -2 < x/2 < 0 => -4 <x <0.

So to compute f(2x) and f(x/2 +2) we need 0<x<1 AND -4<x<0. That means, as you said, x=0

Okay, thank you!