The function $$ f(x)=\frac{4 x+7}{x-2} $$ is one-to-one. a. Find an equation for $$ f^{-1}(x) $$, the inverse function. b. Verify that your equation is correct by showing that $$ f\left(f^{-1}(x)\right)=x $$ and $$ f^{-1}(f(x))=x $$. (Simplify your answers. Use integers or fractions for any numbers in the expression.) $$ \begin{array}{rlrlr} f\left(f^{-1}(x)\right) & =f(x) & \text { and } & f^{-1}(f(x)) & =f^{-1}(x) \ & =x & & \text { Substitute. } \ & =x & & \text { Simplify. } \end{array} $$ The equation is verified.

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```markdown
We are given 
$$
f(x)=\frac{4x+7}{x-2},
$$
and we need to find its inverse $$ f^{-1}(x) $$ and verify that the compositions $$ f\left(f^{-1}(x)\right) $$ and $$ f^{-1}(f(x)) $$ yield $$ x $$.

### (a) Finding the Inverse Function

1. **Write the function in terms of $$ y $$:**
   $$
   y = \frac{4x+7}{x-2}.
   $$

2. **Interchange $$ x $$ and $$ y $$ to solve for $$ y $$:**
   $$
   x = \frac{4y+7}{y-2}.
   $$

3. **Clear the fraction by multiplying both sides by $$ y-2 $$:**
   $$
   x(y-2) = 4y+7.
   $$

4. **Distribute $$ x $$:**
   $$
   xy - 2x = 4y + 7.
   $$

5. **Gather like terms (group the terms with $$ y $$ on one side):**
   $$
   xy - 4y = 2x + 7.
   $$

6. **Factor $$ y $$ from the left-hand side:**
   $$
   y(x-4) = 2x + 7.
   $$

7. **Solve for $$ y $$:**
   $$
   y=\frac{2x+7}{x-4}.
   $$

8. **Rewrite using the inverse function notation:**
   $$
   f^{-1}(x)=\frac{2x+7}{x-4}.
   $$

### (b) Verifying the Inverse Function

We now verify by showing both $$ f\left(f^{-1}(x)\right)=x $$ and $$ f^{-1}(f(x))=x $$.

#### 1. Verify $$ f\left(f^{-1}(x)\right)=x $$

- **Substitute $$ f^{-1}(x) $$ into $$ f $$:**
  $$
  f\left(f^{-1}(x)\right)=f\left(\frac{2x+7}{x-4}\right)=\frac{4\left(\frac{2x+7}{x-4}\right)+7}{\frac{2x+7}{x-4}-2}.
  $$

- **Simplify the numerator:**
  $$
  4\left(\frac{2x+7}{x-4}\right)+7=\frac{8x+28}{x-4}+\frac{7(x-4)}{x-4}=\frac{8x+28+7x-28}{x-4}=\frac{15x}{x-4}.
  $$

- **Simplify the denominator:**
  $$
  \frac{2x+7}{x-4}-2=\frac{2x+7-2(x-4)}{x-4}=\frac{2x+7-2x+8}{x-4}=\frac{15}{x-4}.
  $$

- **Combine the results:**
  $$
  f\left(f^{-1}(x)\right)=\frac{\frac{15x}{x-4}}{\frac{15}{x-4}}=x.
  $$

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

- **Substitute $$ f(x) $$ into $$ f^{-1} $$:**
  $$
  f^{-1}(f(x))=f^{-1}\left(\frac{4x+7}{x-2}\right)=\frac{2\left(\frac{4x+7}{x-2}\right)+7}{\frac{4x+7}{x-2}-4}.
  $$

- **Simplify the numerator:**
  $$
  2\left(\frac{4x+7}{x-2}\right)+7=\frac{8x+14}{x-2}+\frac{7(x-2)}{x-2}=\frac{8x+14+7x-14}{x-2}=\frac{15x}{x-2}.
  $$

- **Simplify the denominator:**
  $$
  \frac{4x+7}{x-2}-4=\frac{4x+7-4(x-2)}{x-2}=\frac{4x+7-4x+8}{x-2}=\frac{15}{x-2}.
  $$

- **Combine the results:**
  $$
  f^{-1}(f(x))=\frac{\frac{15x}{x-2}}{\frac{15}{x-2}}=x.
  $$

### Conclusion

The inverse function is correctly found and verified:
$$
f^{-1}(x)=\frac{2x+7}{x-4},
$$
and we have demonstrated that
$$
f\left(f^{-1}(x)\right)=x \quad \text{and} \quad f^{-1}(f(x))=x.
$$
```
The inverse function is $$ f^{-1}(x) = \frac{2x + 7}{x - 4} $$. This inverse is verified by showing that $$ f(f^{-1}(x)) = x $$ and $$ f^{-1}(f(x)) = x $$.

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