## Discussion

If ((1 – x)*)* = (1 – x)*, then x =
 (A) 1/2 (B) ... (C) ... (D) ... (E) ... (F) ...
*This question is included in Nova Math - Diagnostic/Review:, question #30

The solution is

Posted: 03/21/2012 00:40
Why did the * in the second step disappear and a 1 appear ?
Posted: 04/07/2012 14:59
Yes, don't understand either....why is 1 being substituted for *
Posted: 04/07/2012 19:05
This is the strangest thing I have ever seen and the explanation is not very clear I would very much like a better explanation.
Posted: 04/07/2012 23:19
Hi guys, in "defined functions" problem like this, think of any symbol like * here that defines a function as ƒ(..). So, think of x* as ƒ(x), which in this case, ƒ(x) = 1-x

Let's work on the left hand side of the equation first.
(1-x)* = ƒ(1-x) = 1 - (1-x) = 1 - 1 + x = x
So, ((1-x)*)* = (x)* = ƒ(x) = 1-x.

Now, let's work on the right side. But we already did that. (1-x)* = x.

Hence, we have: 1-x = x, and so x = ....

I hope that helps.
Posted: 04/15/2012 07:51
It really helps, thanks!
Posted: 02/01/2017 09:38
Posted: 04/16/2012 00:18
Thank you Vicki for the acknowledgment. If you have a moment, please rate the app in the App Store. We appreciate it.
Posted: 05/16/2012 18:01
How does (1-x)*= 1-(1-x)?
Posted: 05/16/2012 18:30
Tiffany, thanks for your question. I explained this pretty thoroughly (please look at earlier / above explanations). To repeat:

x* or ƒ(x) is defined as 1 - x. So, you substitute any variable, whether it's x, n, 1, a, M+1, y-5, etc. that has the * into the x in 1 - x.

In this case, for (1-x)* or ƒ(1-x), plug in 1-x into the x of the function.

So you have 1 - (1 - x).

Posted: 05/16/2012 18:35
Oh I see it now thanks very much!
Posted: 05/16/2012 19:00
Tiffany, as Grandpa Oei would have said: kam sia lu, for the acknowledgment. If you have a moment, please rate the app in the App Store.
Posted: 11/30/2012 09:32
If x* = 1 - x
And
((1 - x)*)* = (1 - x)*
Then
((x*)*)* = (x*)*
x^3* = x^2*
x* = 1
(x*)^(1/*) = 1^(1/*)
x = 1

The statement defining the problem defines the symbol *. It does not define the function. Therefore, equating x* to f(x) is incorrect. Perhaps that is intended, but nonetheless it is incorrect.

Posted: 11/30/2012 12:54
Mike, you can't just raise x to the power of 3 or 2 just because x is operated on by the symbol * 3 or 2 times. Trust me, it is easier to think of it like f(x). If you don't understand it then you need to review the chapter on functions.
Posted: 01/02/2013 02:51
With defined function, why don't we distribute the symbol * as we'd do with other variables. Seen on the left side equation (1-x)* becomes (1- x*) which equals, (1 - (1-x). But why don't we distribute the * to the 1 so it becomes, ( *-x*). This is we're it becomes unclear.
Posted: 01/26/2013 13:12
How (1-x)* can be equal to 1-x*?
Posted: 04/12/2013 02:16
Love the app
Posted: 04/16/2013 19:18
Hello Mac. We love the students.
Posted: 11/13/2013 20:34
This is beyond confusing. I have no idea where to start.
Posted: 11/14/2013 17:03
Scott, Joel Brainer explained this in the thread. Here I repaste his answer:

Hi guys, in "defined functions" problem like this, think of any symbol like * here that defines a function as ƒ(..). So, think of x* as ƒ(x), which in this case, ƒ(x) = 1-x

Let's work on the left hand side of the equation first.
(1-x)* = ƒ(1-x) = 1 - (1-x) = 1 - 1 + x = x

So, ((1-x)*)* = (x)* = ƒ(x) = 1-x. (Substitute x into (1-x)* since we just established that.)

Now, let's work on the right side. But we also did that. (1-x)* = x.

Hence, we have: 1-x = x, and so x = ....
Posted: 12/08/2013 22:20
I think 1^* is undefined.
Posted: 12/16/2013 17:00
Mohammed, sorry for the confusion. There is no exponential or power function in the statement. There is no 1^*. x* is defined as 1-x. So if you meant 1*, it is 1-1 or 0.
Posted: 01/22/2014 07:29
But 1* would not be 1-1 if * = x-1.
And if * = x-1, then wouldn't RHS be: (x-1)(x-1)(x-1)?
| Edit
Posted: 03/27/2014 16:48
Cali, the definition is 1-x, not x-1
| Edit
Posted: 10/13/2016 08:20
(1-x)^*=[1-(1-x)]^*???
So (1-x)^2=1-x^2????? You need to be signed in to perform that action.