Prove that the function` f : R to R ` defined by ` f(x)= 2x+ 7 ` , is invertible. Hence find ` f^(-1) `

Answer

To prove that the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = 2x + 7 \) is invertible, we need to show that it is both one-to-one (injective) and onto (surjective). After establishing that the function is invertible, we will find its inverse.

### Step 1: Check if the function is one-to-one
To check if \( f \) is one-to-one, we assume that \( f(x_1) = f(x_2) \) for some \( x_1, x_2 \in \mathbb{R} \).

\[ f(x_1) = f(x_2) \implies 2x_1 + 7 = 2x_2 + 7 \]

Subtracting 7 from both sides:

\[ 2x_1 = 2x_2 \]

Dividing both sides by 2:

\[ x_1 = x_2 \]

Since \( x_1 = x_2 \), this shows that \( f \) is one-to-one.

### Step 2: Check if the function is onto
To check if \( f \) is onto, we need to show that for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \).

Let \( y = f(x) \):

\[ y = 2x + 7 \]

Rearranging this equation to solve for \( x \):

\[ 2x = y - 7 \]
\[ x = \frac{y - 7}{2} \]

Since \( y \) can be any real number, we can find a corresponding \( x \) for every \( y \). Thus, \( f \) is onto.

### Conclusion on invertibility
Since \( f \) is both one-to-one and onto, we conclude that \( f \) is invertible.

### Step 3: Find the inverse function \( f^{-1} \)
From our rearrangement in the onto check, we found:

\[ x = \frac{y - 7}{2} \]

To express this in terms of \( f^{-1}(y) \):

\[ f^{-1}(y) = \frac{y - 7}{2} \]

To express the inverse function in terms of \( x \):

\[ f^{-1}(x) = \frac{x - 7}{2} \]

### Final Result
Thus, the inverse function is:

\[ f^{-1}(x) = \frac{x - 7}{2} \]