this is not correct, so i guess i do not understand.solution 1 = -1
solution 2 = 13/2
final solution [-1,13/2]
this is not correct, so i guess i do not understand.solution 1 = -1
solution 2 = 13/2
final solution [-1,13/2]
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.this is not correct, so i guess i do not understand.
that was incorrect #15 is correctIn 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.
I disagree: both [imath]2x[/imath] and [imath]\frac{x}{2}+2[/imath] must be in [imath][0,2][/imath]that was incorrect #15 is correct
We've all but told you the answer to this. There is a simple fact you aren't remembering.solution 1 = -1
solution 2 = 13/2
final solution [-1,13/2]
hmWe'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
Let's try this. For what values of x is (x/2) + 2 in [0, 2]?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.
I am sorry but what I wrote above is complete nonsense. Sorry about that.I'll do the 1st one for you.
0<x<2. Now double all three terms. This gives you 0<2x<4
Actually, it was incorrect. Sorry.that was incorrect #15 is correct
Okay, thank you!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