If `'f'` is an increasing function from `RvecR` such that `f^(x)>0a n df^(-1)` exists then `(d^2(f^(-1)(x)))/(dx^2)` is `<0` b. `>0` c. `=0` d. cannot be determined

Options

A. lt 0
B. gt 0
C. 0
D. cannot be determined

Explanation

To solve the problem, we need to analyze the properties of the increasing function \( f \) and its inverse \( f^{-1} \). We will derive the second derivative of the inverse function step by step.

### Step 1: Understand the properties of \( f \)
Given that \( f \) is an increasing function and \( f'(x) > 0 \), we know that the first derivative of \( f \) is always positive. This implies that \( f \) is strictly increasing.

**Hint:** Recall that if a function is increasing, its derivative is positive.

### Step 2: Define the inverse function
Let \( g(x) = f^{-1}(x) \). By the definition of the inverse function, we have:
\[ f(g(x)) = x \]

**Hint:** Remember that the inverse function undoes the original function.

### Step 3: Differentiate both sides
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}[f(g(x))] = \frac{d}{dx}[x] \]
Using the chain rule on the left side:
\[ f'(g(x)) \cdot g'(x) = 1 \]
Thus, we can express \( g'(x) \):
\[ g'(x) = \frac{1}{f'(g(x))} \]

**Hint:** Use the chain rule when differentiating composite functions.

### Step 4: Analyze the first derivative
Since \( f'(x) > 0 \), it follows that \( g'(x) > 0 \) as well, because the reciprocal of a positive number is also positive.

**Hint:** The positivity of \( f' \) ensures that \( g' \) is also positive.

### Step 5: Differentiate again to find the second derivative
Now we differentiate \( g'(x) \) to find \( g''(x) \):
\[ g''(x) = \frac{d}{dx}\left(\frac{1}{f'(g(x))}\right) \]
Using the quotient rule:
\[ g''(x) = -\frac{f''(g(x)) \cdot g'(x)}{(f'(g(x)))^2} \]

**Hint:** Remember to apply the quotient rule when differentiating a fraction.

### Step 6: Analyze the second derivative
Since \( f''(x) \) is not specified, we need to analyze the sign of \( g''(x) \):
- \( f'(g(x)) > 0 \) (since \( f \) is increasing)
- \( g'(x) > 0 \) (as derived earlier)
- The sign of \( f''(g(x)) \) will determine the sign of \( g''(x) \).

However, since \( f \) is increasing and \( f' > 0 \), if \( f''(x) < 0 \) (which is a common case for concave functions), then \( g''(x) < 0 \).

**Hint:** The sign of the second derivative can indicate concavity or convexity.

### Conclusion
Thus, if \( f''(x) < 0 \), we conclude that:
\[ g''(x) < 0 \]
This implies that \( \frac{d^2}{dx^2} f^{-1}(x) < 0 \).

### Final Answer
The answer is:
**(a) < 0**