Thread: given f(x) and g(x) i am asked to show an equality

f(x)= 3/sqrt(9+4x) and g(x)= 3/sqrt(1+4x)

i am asked to show that by writing

3/sqrt(9+4x)= 1/sqrt(1+(4x/9)) (and here their is some factor that i know is there, obs) but i cannot see.)

and am then asked to substitute into one of the standard taylor series about 0 for f.give the explicity upto the term x^3 and determine the range of validity.My algebra is letting me down again.I know what i want but i dont know how to get it.

Can anyone help????

Many thanks in advance....

They have written 9+4x as 9(1+4x/9) so the square root is 3root(1+4x/9) and then the 3 has cancelled out.

confused........

if i write (3/sqrt(9+4x))= ?*((1/sqrt(1+(4/9x)) and solve for ? i get the value of 1.this would make sense but i am missing something.

The square root of a product= the product of the separate square roots. Hence square root of 9(1+4/9x)=3sqrt(1+4/9x)