Math Is Fun Forum

Indeed, this is the first step in designing every part of a new application.

The main, if not crucial, role of the human scientific brain is to find out the formula, the function or the equation which can emulate/reflect, as possible, a real problem that needs to be solved.

Therefore, the purpose of the various math's exercises is to gain, also as possible, the logical reasoning on how to get the required/needed results from analyzing a formula/function of interest or solving a well-defined equation(s). Fortunately, there are also many ready-made tools to assist someone in doing this to save time (after he learns how to use them).

The calculus way is based on the knowledge of the derivatives of functions.

Thank you, Bob.
Your steps are much better than mine.

It seems you did it.
But on my side, I just see the file name of a hidden image which is 'IbFoc46.jpeg'.

I did it by following primitive steps since I forgot, at age 75, the advanced ones.

f(n)   = f(n-1)*2+1				
f(n-1) = f(n-2)*2+1				
	f(n)   = (f(n-2)*2+1)*2+1			
	f(n-2) = f(n-3)*2+1			
		f(n)   = ((f(n-3)*2+1)*2+1)*2+1		
		f(n-3) = f(n-4)*2+1		
			f(n)   = (((f(n-4)*2+1)*2+1)*2+1)*2+1	
			f(n-4) = f(n-5)*2+1	
				f(n) = ((((f(n-5)*2+1)*2+1)*2+1)*2+1)*2+1
				
f(n) = ((((f(n-5)*2+1)*2+1)*2+1)*2+1)*2+1				
f(n) = (((f(n-5)*2+1)*2+1)*2+1)*2*2+2+1				
f(n) = (((f(n-5)*2+1)*2+1)*2*2*2+2*2+2+1				
f(n) = (((f(n-5)*2+1)*2*2*2*2+2*2*2+2*2+2+1				
f(n) = f(n-5)*2*2*2*2*2+2*2*2*2+2*2*2+2*2+2+1				
f(n) = f(n-a)*2^a… +2^(a-1)+2^(a-2)+2^(a-3)+2^(a-4)+1				

Um=k*r^(m-1)				
In the following series
… +2^(a-1)+2^(a-2)+2^(a-3)+2^(a-4)+1
we have
k=1
m=a			
f(n) = f(n-a)*2^a+2^(a)-1				
f(n) = 2^a*[f(n-a)+1]-1				

Let us assume:
n-a=1				
n=a+1
f(a+1) = 2^a*[f(1)+1]-1				
But
f(1)= 2				
Therefore
f(a+1) = 2^a*3-1				

Again, let us assume:
a+1=n				
a=n-1				
f(n) = 2^(n-1)*3-1				
f(n) = 3*2^(n-1)-1				

What confuses me is how one gives a result as 'a range' and as 'a domain'!

"Given A(x) = 4x•sqrt{1 - x^2}, find the 'domain' of A."

A(x) is supposed to be a dependent variable, so I expected to read... find the 'range' of A.

I suggest, till you will succeed doing it on your side, that you may like to include the URL of your image of interest with a posted question and the first member who will read your post will attach it to a subsequent post for you (as I did here).

What do you think?

Hi Bob,
It seems that that 'range and 'domain' could be applied both on the variable or the function.
I thought, so I may be wrong, that 'domain' is for the variable only and 'range' is for the function only.
Please clarify this point.
Thank you.
Kerim

I expected to read instead:
B. Find domain of x algebraically

The best, fastest and simplest tactic to gain 'huge' amount of money from the ordinary people/multitudes of a country is ... (fill the gap)

Added:
Hint: It doesn't work with me in the way I live.