Use the following table and the fact that `f (x)` is invertible and differentiable everywhere to find `(f ^(-1)(3))':`
`{:(x, f (x), f '(x)),(3,1,7),(6,2,10),(9,3,5):}`

Options

A. 0
B. `1/5`
C. `1/10`
D. `1/7`

Answer

To find \((f^{-1})'(3)\) using the given table, we will follow these steps:

### Step 1: Understand the relationship between \(f\) and \(f^{-1}\)

We know that if \(y = f(x)\), then \(x = f^{-1}(y)\). The derivative of the inverse function can be expressed as:
\[ (f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))} \]

### Step 2: Identify \(f^{-1}(3)\)

From the table provided, we need to find \(x\) such that \(f(x) = 3\). Looking at the table:

\[ \begin{array}{|c|c|c|} \hline x & f(x) & f'(x) \\ \hline 3 & 1 & 7 \\ 6 & 2 & 10 \\ 9 & 3 & 5 \\ \hline \end{array} \]

We see that when \(x = 9\), \(f(9) = 3\). Therefore, we have:
\[ f^{-1}(3) = 9 \]

### Step 3: Find \(f'(f^{-1}(3))\)

Now, we need to find \(f'(f^{-1}(3))\). Since \(f^{-1}(3) = 9\), we need to find \(f'(9)\) from the table.

From the table, we see that:
\[ f'(9) = 5 \]

### Step 4: Calculate \((f^{-1})'(3)\)

Using the formula for the derivative of the inverse function:
\[ (f^{-1})'(3) = \frac{1}{f'(f^{-1}(3))} = \frac{1}{f'(9)} = \frac{1}{5} \]

### Final Answer

Thus, \((f^{-1})'(3) = \frac{1}{5}\).

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