Suppose f,g, `R to R`. If g(x) defined by `g(x) = x^(2) + x-2` and (g.f)(x) `=4x^(2) - 10 x + 4`, then `f(x)` may be given by

Options

A. `f(x) = 2x+3`
B. `f(x) = 3-2x`
C. `f(x) = 2x-3`
D. `f(x) = 2+2x`

Answer

To find the function \( f(x) \) given the function \( g(x) = x^2 + x - 2 \) and the composition \( (g \circ f)(x) = 4x^2 - 10x + 4 \), we will follow these steps:

### Step 1: Write the expression for \( g(f(x)) \)

Since \( g(x) = x^2 + x - 2 \), we can substitute \( f(x) \) into \( g \):

\[ g(f(x)) = (f(x))^2 + f(x) - 2 \]

### Step 2: Set up the equation

We know that \( g(f(x)) = 4x^2 - 10x + 4 \). Therefore, we can set up the equation:

\[ (f(x))^2 + f(x) - 2 = 4x^2 - 10x + 4 \]

### Step 3: Rearrange the equation

Rearranging the equation gives us:

\[ (f(x))^2 + f(x) - 4x^2 + 10x - 6 = 0 \]

### Step 4: Identify coefficients

This is a quadratic equation in terms of \( f(x) \):

\[ (f(x))^2 + f(x) + (-4x^2 + 10x - 6) = 0 \]

Here, we identify:
- \( a = 1 \)
- \( b = 1 \)
- \( c = -4x^2 + 10x - 6 \)

### Step 5: Use the quadratic formula

We can apply the quadratic formula \( f(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ f(x) = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-4x^2 + 10x - 6)}}{2 \cdot 1} \]

### Step 6: Simplify under the square root

Calculating the discriminant:

\[ 1 + 16x^2 - 40x + 24 = 16x^2 - 40x + 25 \]

### Step 7: Factor the discriminant

The expression \( 16x^2 - 40x + 25 \) can be factored as:

\[ (4x - 5)^2 \]

### Step 8: Substitute back into the formula

Now substituting back into the quadratic formula:

\[ f(x) = \frac{-1 \pm (4x - 5)}{2} \]

### Step 9: Solve for \( f(x) \)

Calculating both cases:

1. Taking the positive root:
\[ f(x) = \frac{-1 + (4x - 5)}{2} = \frac{4x - 6}{2} = 2x - 3 \]

2. Taking the negative root:
\[ f(x) = \frac{-1 - (4x - 5)}{2} = \frac{-4x + 4}{2} = -2x + 2 \]

### Conclusion

Thus, the function \( f(x) \) can be either:

\[ f(x) = 2x - 3 \quad \text{or} \quad f(x) = -2x + 2 \]