Let `f :[0,2pi] to [-3, 3]` be a given function defined at `f (x) = 3cos ""(x)/(2).` The slope of the tangent to the curve `y = f^(-1) (x)` at the point where the curve crosses the y-axis is:

Options

A. a) `-1`
B. b) `-2/3`
C. c) `-1/6`
D. d) `-1/3`

Answer

To solve the problem, we need to find the slope of the tangent to the curve \( y = f^{-1}(x) \) at the point where it crosses the y-axis. The function given is \( f(x) = 3 \cos\left(\frac{x}{2}\right) \).

### Step-by-Step Solution:

**Step 1: Find the inverse function \( f^{-1}(x) \)**

We start with the equation \( y = f(x) = 3 \cos\left(\frac{x}{2}\right) \). To find the inverse, we set \( z = f(x) \), so:

\[ z = 3 \cos\left(\frac{x}{2}\right) \]

Now, we solve for \( x \):

1. Divide both sides by 3:
\[ \cos\left(\frac{x}{2}\right) = \frac{z}{3} \]

2. Take the inverse cosine:
\[ \frac{x}{2} = \cos^{-1}\left(\frac{z}{3}\right) \]

3. Multiply by 2 to solve for \( x \):
\[ x = 2 \cos^{-1}\left(\frac{z}{3}\right) \]

Thus, the inverse function is:
\[ f^{-1}(x) = 2 \cos^{-1}\left(\frac{x}{3}\right) \]

**Step 2: Find the derivative \( \frac{dy}{dx} \)**

To find the slope of the tangent line to the curve \( y = f^{-1}(x) \), we differentiate \( y \) with respect to \( x \):

\[ y = 2 \cos^{-1}\left(\frac{x}{3}\right) \]

Using the chain rule, we differentiate:

\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx} \left(\cos^{-1}\left(\frac{x}{3}\right)\right) \]

The derivative of \( \cos^{-1}(u) \) is \( -\frac{1}{\sqrt{1 - u^2}} \) where \( u = \frac{x}{3} \). Thus:

\[ \frac{dy}{dx} = 2 \cdot \left(-\frac{1}{\sqrt{1 - \left(\frac{x}{3}\right)^2}}\right) \cdot \frac{1}{3} \]

Simplifying this gives:

\[ \frac{dy}{dx} = -\frac{2}{3\sqrt{1 - \left(\frac{x}{3}\right)^2}} \]

**Step 3: Evaluate the derivative at the y-axis**

The curve crosses the y-axis when \( x = 0 \). We substitute \( x = 0 \) into the derivative:

\[ \frac{dy}{dx}\bigg|_{x=0} = -\frac{2}{3\sqrt{1 - \left(\frac{0}{3}\right)^2}} = -\frac{2}{3\sqrt{1}} = -\frac{2}{3} \]

### Final Answer:

The slope of the tangent to the curve \( y = f^{-1}(x) \) at the point where it crosses the y-axis is:

\[ \boxed{-\frac{2}{3}} \]