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Question: Suppose that \(g(x) = 1 + \sqrt{x}\) and \(f(g(x)) = 3 + 2\sqrt{x} + x,\) then f(x) is...

Suppose that g(x)=1+xg(x) = 1 + \sqrt{x} and f(g(x))=3+2x+x,f(g(x)) = 3 + 2\sqrt{x} + x, then

f(x) is

A

f−1(y)=x+43f^{- 1}(y) = \frac{x + 4}{3}

B

2+x22 + x^{2}

C

1+x1 + x

D

2+x2 + x

Answer

2+x22 + x^{2}

Explanation

Solution

g(x)=1+xg(x) = 1 + \sqrt{x} and f(g(x))=3+2x+xf(g(x)) = 3 + 2\sqrt{x} + x ….. (i)

⇒ ⇒\Rightarrow

Put 1+x=y1 + \sqrt{x} = y ⇒ x=(y−1)2x = (y - 1)^{2}

then, f(y)=3+2(y−1)+(y−1)2=2+y2f(y) = 3 + 2(y - 1) + (y - 1)^{2} = 2 + y^{2}

therefore, f(x)=2+x2f(x) = 2 + x^{2}.