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I reached till \(\frac{1}{(1+x)}\) thats alright.. But how did you guys know you had to subtract this value from 1 to get the answer in terms of \(f(x)\)? _________________

so f(x)= x/x+1. it is asking what f(1/x) is in terms of f(x).

So to find that you plug 1/x in for x in f(x). the result is,

(1/x)/[(1/x)+1]

since you have two fractions you will want to multiply by a reciprocal, but before that lets simply the denominator.

denominator= [(1/x)+1] you can simplify that by adding them together to create one term but to do that the denominators must be the same, so you end up with [(1/x)+(x/x)]. *note that (x/x)=1

now your equation for f(1/x) looks like (1/x)/[(1+x)/x]

multiply by the reciprocal so (1/x)*(x/1+x), the x in the denominator of 1/x cancels out with the x in the numerator of x/(1+x) and you end up with 1/(1+x)

now you have to find out what that equals in terms of f(x).

A) cant be because it is not the same equation of f(x) B) This is not the cause because it is not -(x/(x+1)) C) Hold off on this one because a little algebra needs to be down D)1-f(x)= 1- (x/x+1) which equals ((x+1)/(x+1))-(1/(x+1))---->x/(x+1) which is f(x)

For Algebra questions, I think the easiest way is plug-in. Clearly see f(1/x) = (1/x) / ((1/x) +1) = 1/(x+1). Don't take time to convert 1/(x+1) to a formula in terms of f(x), just examine each answer.

Ans A: f(x) = x/(x+1) wrong Ans B: -f(x) = -x/(x+1) Wrong Ans C: 1/f(x) = (x+1)/x Wrong Ans D: 1 - f(x) = 1 - x/(x+1) = 1/(x+1) CORRECT _________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

The best way to solve these kind of questions is to assume values... For example take x=2...with this value of x you will get f(x) as 2/3... Now find f(1/x) i.e. f(1/2), which will come out to be 1/3.... Now check which option satisfies the above relation...the only option that satisfies is D