If f(x) = \(\frac{1}{(2x+1)}\) and x ≠ \(\frac{-1}2\) then pr

1.	If f(x) = \(\frac{1}{(2x+1)}\) and x ≠ \(\frac{-1}2\) then prove that f{(x)} = \(\frac{2x+1}{2x+3}\), when it is given that x ≠ \(\frac{-3}2\).
Answer» Given: f(x) = \(\frac{1}{(2x+1)}\), where x ≠ \(\frac{-1}2\) Need to prove: f{f(x)} = \(\frac{2x +1}{2x+3}\) When x ≠ \(\frac{-3}2\) Now placing f(x) in place of x ⇒ f{f(x)} = \(\frac{1}{2f(x) + 1}\) ⇒ f{f(x)} = \(\frac{1}{2\frac{1}{2x+1 }+1}\) ⇒ f{f(x)} = \(\frac{1}{\frac{2+2x+1}{2x+1 }}\) = \(\frac{2x +1}{2x+3}\), Where x ≠ \(\frac{-3}2\) [proved]

If f(x) = \(\frac{1}{(2x+1)}\) and x ≠ \(\frac{-1}2\) then prove that f{(x)} = \(\frac{2x+1}{2x+3}\), when it is given that x ≠ \(\frac{-3}2\).

Answer»

Given: f(x) = \(\frac{1}{(2x+1)}\), where x ≠ \(\frac{-1}2\)

Need to prove: f{f(x)} = \(\frac{2x +1}{2x+3}\) When x ≠ \(\frac{-3}2\)

Now placing f(x) in place of x

⇒ f{f(x)} = \(\frac{1}{2f(x) + 1}\)

⇒ f{f(x)} = \(\frac{1}{2\frac{1}{2x+1 }+1}\)

⇒ f{f(x)} = \(\frac{1}{\frac{2+2x+1}{2x+1 }}\) = \(\frac{2x +1}{2x+3}\), Where x ≠ \(\frac{-3}2\)

[proved]