The function $$ f(x)=7 x+5 $$ is one-to-one. a. Find an equation for $$ f^{-1} $$, the inverse function. b. Verify that your equation is correct by showing that $$ f\left(f^{-1}(x)\right)=x $$ and $$ f^{-1}(f(x))=x $$.

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### a. Finding the Inverse Function $$ f^{-1}(x) $$

To find the inverse function $$ f^{-1}(x) $$ of the given function $$ f(x) = 7x + 5 $$, follow these steps:

1. **Start with the original function:**
   $$
   y = 7x + 5
   $$

2. **Swap $$ x $$ and $$ y $$:**
   $$
   x = 7y + 5
   $$

3. **Solve for $$ y $$ to find the inverse:**
   $$
   x - 5 = 7y \
   y = \frac{x - 5}{7}
   $$

4. **Express the inverse function:**
   $$
   f^{-1}(x) = \frac{x - 5}{7}
   $$

### b. Verifying the Inverse Function

To verify that $$ f^{-1}(x) $$ is indeed the inverse of $$ f(x) $$, we need to show that both compositions $$ f(f^{-1}(x)) $$ and $$ f^{-1}(f(x)) $$ simplify to $$ x $$.

#### 1. Verifying $$ f(f^{-1}(x)) = x $$

Start with the inverse function inside $$ f $$:
$$
f(f^{-1}(x)) = f\left(\frac{x - 5}{7}\right)
$$

Apply the original function $$ f $$ to this expression:
$$
f\left(\frac{x - 5}{7}\right) = 7 \left(\frac{x - 5}{7}\right) + 5
$$

Simplify the expression:
$$
= \frac{7(x - 5)}{7} + 5 \
= (x - 5) + 5 \
= x
$$

#### 2. Verifying $$ f^{-1}(f(x)) = x $$

Start with the original function inside the inverse:
$$
f^{-1}(f(x)) = f^{-1}(7x + 5)
$$

Apply the inverse function to this expression:
$$
f^{-1}(7x + 5) = \frac{(7x + 5) - 5}{7}
$$

Simplify the expression:
$$
= \frac{7x}{7} \
= x
$$

### Conclusion

Both compositions $$ f(f^{-1}(x)) $$ and $$ f^{-1}(f(x)) $$ simplify to $$ x $$, confirming that $$ f^{-1}(x) = \frac{x - 5}{7} $$ is indeed the inverse function of $$ f(x) = 7x + 5 $$.
a. The inverse function is $$ f^{-1}(x) = \frac{x - 5}{7} $$. b. Both $$ f(f^{-1}(x)) $$ and $$ f^{-1}(f(x)) $$ equal $$ x $$, confirming that $$ f^{-1}(x) $$ is the correct inverse.

The function \( f(x)=7 x+5 \) is one-to-one. a. Find an equation for \( f^{-1} \), the inverse function. b. Verify that your equation is correct by showing that \( f\left(f^{-1}(x)\right)=x \) and \( f^{-1}(f(x))=x \).

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