If `g(x)=1+sqrtx and f(g(x))=3+2sqrtx+x` then f(x) is equal to

Options

A. `1+2x^(2)`
B. `2+x^(2)`
C. `1+x`
D. 2+x

Answer

To find the function \( f(x) \) given that \( g(x) = 1 + \sqrt{x} \) and \( f(g(x)) = 3 + 2\sqrt{x} + x \), we can follow these steps:

### Step 1: Substitute \( g(x) \) into \( f(g(x)) \)

We know:
\[ g(x) = 1 + \sqrt{x} \]
Thus, we can write:
\[ f(g(x)) = f(1 + \sqrt{x}) = 3 + 2\sqrt{x} + x \]

### Step 2: Rewrite \( f(g(x)) \)

We can rearrange the expression \( 3 + 2\sqrt{x} + x \):
\[ 3 + 2\sqrt{x} + x = 2 + 1 + 2\sqrt{x} + x \]
This can be grouped as:
\[ = 2 + (1 + \sqrt{x})^2 \]
since \( (1 + \sqrt{x})^2 = 1 + 2\sqrt{x} + x \).

### Step 3: Identify \( f(x) \)

From the previous step, we have:
\[ f(1 + \sqrt{x}) = 2 + (1 + \sqrt{x})^2 \]
Now, let \( u = 1 + \sqrt{x} \). Then, we can express \( \sqrt{x} \) as \( u - 1 \) and substitute back:
\[ f(u) = 2 + u^2 \]

### Step 4: Write the final expression for \( f(x) \)

Thus, the function \( f(x) \) can be expressed as:
\[ f(x) = 2 + x^2 \]

### Step 5: Verify the solution

To ensure our solution is correct, we can check if substituting \( g(x) \) back into \( f(x) \) gives us \( f(g(x)) \):
\[ f(g(x)) = f(1 + \sqrt{x}) = 2 + (1 + \sqrt{x})^2 \]
Calculating this gives:
\[ = 2 + (1 + 2\sqrt{x} + x) = 2 + 1 + 2\sqrt{x} + x = 3 + 2\sqrt{x} + x \]
This matches the original expression for \( f(g(x)) \), confirming our solution is correct.

### Final Answer

Thus, the function \( f(x) \) is:
\[ \boxed{2 + x^2} \]