Let `f(x) = 3x + 5 AA x in R, g^(-1)(x) = x^(3) +1 AA x in R`, then `(f^(-1).g)^(-1)(x)` is equal to

Options

A. `(3x + 5)^(3)`
B. `(3x + 5)^(3) +1`
C. `1-(3x + 5)^(3)`
D. none of these

Answer

To solve the problem, we need to find \((f^{-1} \circ g)^{-1}(x)\) given the functions \(f(x) = 3x + 5\) and \(g^{-1}(x) = x^3 + 1\).

### Step-by-Step Solution:

1. **Find \(f^{-1}(x)\)**:
- Start with the equation \(y = f(x) = 3x + 5\).
- To find the inverse, solve for \(x\):
\[ y = 3x + 5 \implies 3x = y - 5 \implies x = \frac{y - 5}{3}. \]
- Thus, the inverse function is:
\[ f^{-1}(x) = \frac{x - 5}{3}. \]

2. **Find \(g(x)\)**:
- Since we have \(g^{-1}(x) = x^3 + 1\), we need to find \(g(x)\).
- To find \(g(x)\), we set \(y = g^{-1}(x)\) and solve for \(x\):
\[ x = g(y) \implies y = x^3 + 1 \implies g(y) = x - 1. \]
- Thus, the function \(g(x)\) is:
\[ g(x) = \sqrt[3]{x - 1}. \]

3. **Find \(f^{-1}(g(x))\)**:
- Substitute \(g(x)\) into \(f^{-1}(x)\):
\[ f^{-1}(g(x)) = f^{-1}(\sqrt[3]{x - 1}). \]
- Using the expression for \(f^{-1}(x)\):
\[ f^{-1}(g(x)) = \frac{\sqrt[3]{x - 1} - 5}{3}. \]

4. **Find \((f^{-1} \circ g)^{-1}(x)\)**:
- Let \(y = f^{-1}(g(x))\):
\[ y = \frac{\sqrt[3]{x - 1} - 5}{3}. \]
- To find the inverse, solve for \(x\):
\[ 3y = \sqrt[3]{x - 1} - 5 \implies \sqrt[3]{x - 1} = 3y + 5. \]
- Cubing both sides:
\[ x - 1 = (3y + 5)^3 \implies x = (3y + 5)^3 + 1. \]
- Thus, the inverse function is:
\[ (f^{-1} \circ g)^{-1}(x) = (3x + 5)^3 + 1. \]

### Final Result:
\[ (f^{-1} \circ g)^{-1}(x) = (3x + 5)^3 + 1. \]