The function $$ f(x)=\frac{2}{x} $$ 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 $$ f\left(f^{-1}(x) ight)=x $$ and $$ f^{-1}(f(x))=x $$. B. $$ f^{\prime}(x)=\frac{-}{x} $$, for $$ x
eq 0 $$ D. $$ f^{-1}(x)= $$, for all $$ x $$ b. Verify $$ t $$ that the equation is correct. $$ \begin{array}{l}f\left(f^{-1}(x) ight)=f(\square) \ \ =\square\end{array} $$

Question

The function $$ f(x)=\frac{2}{x} $$ 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 $$ f\left(f^{-1}(x)ight)=x $$ and $$ f^{-1}(f(x))=x $$. B. $$ f^{\prime}(x)=\frac{-}{x} $$, for $$ x 
eq 0 $$ D. $$ f^{-1}(x)= $$, for all $$ x $$ b. Verify $$ t $$ that the equation is correct. $$ \begin{array}{l}f\left(f^{-1}(x)ight)=f(\square) \ \ =\square\end{array} $$

UpStudy AI · Accepted Answer

Certainly! Let's work through each part of your problem step by step.

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

Given the function:
$$ f(x) = \frac{2}{x} $$

To find the inverse function $$ f^{-1}(x) $$, follow these steps:

1. **Replace $$ f(x) $$ with $$ y $$:**
   $$ y = \frac{2}{x} $$

2. **Swap $$ x $$ and $$ y $$ to begin finding the inverse:**
   $$ x = \frac{2}{y} $$

3. **Solve for $$ y $$:**
   $$
   \begin{align*}
   x &= \frac{2}{y} \
   xy &= 2 \quad (\text{Multiply both sides by } y) \
   y &= \frac{2}{x}
   \end{align*}
   $$

4. **Express the inverse function:**
   $$ f^{-1}(x) = \frac{2}{x} $$

**Conclusion:** The inverse function is:
$$ f^{-1}(x) = \frac{2}{x} $$

### **b. Verifying the Inverse Function**

To verify that $$ f^{-1}(x) $$ is indeed the inverse of $$ f(x) $$, we need to show that:
1. $$ f(f^{-1}(x)) = x $$
2. $$ f^{-1}(f(x)) = x $$

**1. Verify $$ f(f^{-1}(x)) = x $$:**
$$
\begin{align*}
f(f^{-1}(x)) &= f\left(\frac{2}{x}\right) \
&= \frac{2}{\frac{2}{x}} \
&= \frac{2 \cdot x}{2} \quad (\text{Simplify the complex fraction}) \
&= x
\end{align*}
$$

**2. Verify $$ f^{-1}(f(x)) = x $$:**
$$
\begin{align*}
f^{-1}(f(x)) &= f^{-1}\left(\frac{2}{x}\right) \
&= \frac{2}{\frac{2}{x}} \
&= \frac{2 \cdot x}{2} \quad (\text{Simplify the complex fraction}) \
&= x
\end{align*}
$$

**Conclusion:** Both compositions yield $$ x $$, confirming that $$ f^{-1}(x) = \frac{2}{x} $$ is indeed the inverse of $$ f(x) = \frac{2}{x} $$.

### **Additional Information**

It seems there were additional parts mentioned in your question. Let's address them for completeness.

#### **B. Finding the Derivative $$ f'(x) $$**

Given:
$$ f(x) = \frac{2}{x} $$

To find the derivative $$ f'(x) $$:
$$
f'(x) = \frac{d}{dx}\left(\frac{2}{x}\right) = -\frac{2}{x^2}
$$

**Conclusion:**
$$ f'(x) = -\frac{2}{x^2} \quad \text{for } x \neq 0 $$

#### **D. Confirming the Inverse Function**

As previously determined:
$$ f^{-1}(x) = \frac{2}{x} $$

#### **Verification Template Completion**

You provided a template to verify the inverse function. Let's fill in the blanks:

$$
\begin{array}{l}
f\left(f^{-1}(x)\right) = f\left(\frac{2}{x}\right) \
= \frac{2}{\frac{2}{x}} \
= x
\end{array}
$$

Similarly:

$$
\begin{array}{l}
f^{-1}(f(x)) = f^{-1}\left(\frac{2}{x}\right) \
= \frac{2}{\frac{2}{x}} \
= x
\end{array}
$$

**Final Summary:**

- **Inverse Function:** $$ f^{-1}(x) = \frac{2}{x} $$
- **Derivative:** $$ f'(x) = -\frac{2}{x^2} $$ for $$ x \neq 0 $$
- **Verification:** Both $$ f(f^{-1}(x)) $$ and $$ f^{-1}(f(x)) $$ simplify to $$ x $$, confirming the correctness of the inverse function.

If you have any further questions or need additional clarifications, feel free to ask!
The inverse function is $$ f^{-1}(x) = \frac{2}{x} $$. To verify, both $$ f(f^{-1}(x)) $$ and $$ f^{-1}(f(x)) $$ equal $$ x $$.

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