# C1 QuestionWatch

#1
So I've been doing some C1 past papers and there's part of a question I can't do.

f(x) = (3-4sqrx)^2
........ -----------
........ sqrx

we have to show f(x) = 9x^-1/2 + Ax^1/2 + B
where A and B are constants to be found.

I've found B and the begininng bit but I am having trouble proving
16x
---- = 16x^1/2
sqrx

0
8 years ago
#2
16x/sqrtx=16x(x^(-1/2))=16x^(1-1/2)=16x^(1/2)
0
8 years ago
#3
(Original post by heymynameisben)
So I've been doing some C1 past papers and there's part of a question I can't do.

f(x) = (3-4sqrx)^2
........ -----------
........ sqrx

we have to show f(x) = 9x^-1/2 + Ax^1/2 + B
where A and B are constants to be found.

I've found B and the begininng bit but I am having trouble proving
16x
---- = 16x^1/2
sqrx

Indices rules?

x^n / x^m = x^(n-m)
0
#4
Thanks a lot!
0
8 years ago
#5
16x
---- = 16x / x^(1/2) When you divide, you minus powers. So its 16x^(1) / x^(1/2) => 1-(1/2) = 1/2. So you get 16x^(1/2)
sqrx
0
8 years ago
#6
Think back to chapter 1 indices. Sqrx = x^1/2, and dividing indices means taking away. 1 - 1/2 is 1/2
0
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