Let `f(x)=x^(3)+x+1` and let g(x) be its inverse function then equation of the tangent to `y=g(x)` at x = 3 is

Options

A. `x-4y+1=0`
B. `x+4y-1=0`
C. `4x-y+1=0`
D. `4x+y-1=0`

Explanation

To find the equation of the tangent to the curve \( y = g(x) \) at \( x = 3 \), where \( g(x) \) is the inverse of \( f(x) = x^3 + x + 1 \), we will follow these steps:

### Step 1: Find the value of \( \alpha \) such that \( g(3) = \alpha \)
Since \( g(x) \) is the inverse of \( f(x) \), we have:
\[ g(3) = \alpha \implies f(\alpha) = 3 \]
This means we need to solve the equation:
\[ \alpha^3 + \alpha + 1 = 3 \]
which simplifies to:
\[ \alpha^3 + \alpha - 2 = 0 \]

### Step 2: Solve the cubic equation
We will use trial and error or synthetic division to find the roots of the cubic equation \( \alpha^3 + \alpha - 2 = 0 \). Testing \( \alpha = 1 \):
\[ 1^3 + 1 - 2 = 0 \]
Thus, \( \alpha = 1 \) is a root.

### Step 3: Factor the cubic equation
Now we can factor \( \alpha^3 + \alpha - 2 \) using \( (\alpha - 1) \):
\[ \alpha^3 + \alpha - 2 = (\alpha - 1)(\alpha^2 + \alpha + 2) \]
The quadratic \( \alpha^2 + \alpha + 2 \) has a negative discriminant (\( 1 - 8 < 0 \)), indicating no real roots. Therefore, the only real solution is \( \alpha = 1 \).

### Step 4: Determine the point on \( g(x) \)
Thus, we have:
\[ g(3) = 1 \implies \text{The point is } (3, 1) \]

### Step 5: Find the derivative \( g'(x) \)
Using the relationship between the derivatives of inverse functions:
\[ g'(x) = \frac{1}{f'(g(x))} \]
At \( x = 3 \):
\[ g'(3) = \frac{1}{f'(g(3))} = \frac{1}{f'(1)} \]

### Step 6: Calculate \( f'(x) \)
First, we find \( f'(x) \):
\[ f'(x) = 3x^2 + 1 \]
Now, substituting \( x = 1 \):
\[ f'(1) = 3(1)^2 + 1 = 4 \]
Thus,
\[ g'(3) = \frac{1}{4} \]

### Step 7: Write the equation of the tangent line
The equation of the tangent line at the point \( (3, 1) \) can be written in point-slope form:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (3, 1) \) and \( m = \frac{1}{4} \):
\[ y - 1 = \frac{1}{4}(x - 3) \]
Rearranging gives:
\[ y = \frac{1}{4}x + \frac{1}{4} \]

### Step 8: Convert to standard form
To convert to standard form, we can multiply through by 4:
\[ 4y = x + 1 \implies x - 4y + 1 = 0 \]

### Final Answer
The equation of the tangent to \( y = g(x) \) at \( x = 3 \) is:
\[ x - 4y + 1 = 0 \]