If the function `f: R rarr R` be defined by `f(x)=2x-3` and `g:R rarr R` by `g(x)=x^(3)+5`, then show that fog is invertible. Also find `(fog)^(-1)(x)`, hence find `(fog)^(-1)(9)`.

Answer

To solve the problem, we need to show that the composition of the functions \( f \) and \( g \) is invertible and then find the inverse function \( (f \circ g)^{-1}(x) \) and evaluate \( (f \circ g)^{-1}(9) \).

### Step 1: Define the Functions
We have:
- \( f(x) = 2x - 3 \)
- \( g(x) = x^3 + 5 \)

### Step 2: Find the Composition \( f \circ g \)
The composition \( f \circ g \) is defined as:
\[ (f \circ g)(x) = f(g(x)) = f(x^3 + 5) \]
Substituting \( g(x) \) into \( f(x) \):
\[ f(x^3 + 5) = 2(x^3 + 5) - 3 = 2x^3 + 10 - 3 = 2x^3 + 7 \]
Thus, we have:
\[ (f \circ g)(x) = 2x^3 + 7 \]

### Step 3: Check if \( f \circ g \) is Invertible
To show that \( f \circ g \) is invertible, we need to check if it is a bijective function (both one-to-one and onto).

#### Step 3.1: Check if \( f \circ g \) is One-to-One
A function is one-to-one if \( f(a) = f(b) \) implies \( a = b \).

Assume:
\[ f \circ g(a) = f \circ g(b) \]
This gives:
\[ 2a^3 + 7 = 2b^3 + 7 \]
Subtracting 7 from both sides:
\[ 2a^3 = 2b^3 \]
Dividing by 2:
\[ a^3 = b^3 \]
Taking the cube root:
\[ a = b \]
Thus, \( f \circ g \) is one-to-one.

#### Step 3.2: Check if \( f \circ g \) is Onto
A function is onto if for every \( y \) in the codomain, there exists an \( x \) in the domain such that \( f \circ g(x) = y \).

Let \( y = 2x^3 + 7 \). Rearranging gives:
\[ 2x^3 = y - 7 \]
\[ x^3 = \frac{y - 7}{2} \]
\[ x = \sqrt[3]{\frac{y - 7}{2}} \]
Since \( y \) can take any real value, we can find a corresponding \( x \) for every \( y \). Therefore, \( f \circ g \) is onto.

### Conclusion
Since \( f \circ g \) is both one-to-one and onto, it is bijective and thus invertible.

### Step 4: Find the Inverse Function \( (f \circ g)^{-1}(x) \)
From our earlier work, we have:
\[ y = 2x^3 + 7 \]
To find the inverse, we solve for \( x \):
\[ y - 7 = 2x^3 \]
\[ x^3 = \frac{y - 7}{2} \]
\[ x = \sqrt[3]{\frac{y - 7}{2}} \]
Thus, the inverse function is:
\[ (f \circ g)^{-1}(x) = \sqrt[3]{\frac{x - 7}{2}} \]

### Step 5: Evaluate \( (f \circ g)^{-1}(9) \)
Now, we substitute \( x = 9 \) into the inverse function:
\[ (f \circ g)^{-1}(9) = \sqrt[3]{\frac{9 - 7}{2}} = \sqrt[3]{\frac{2}{2}} = \sqrt[3]{1} = 1 \]

### Final Answer
Thus, \( (f \circ g)^{-1}(9) = 1 \).