Consider `f(x)=(x)/(x^(2)-1)`
Statement I Graph of `f(x)` is concave up for `xgt1.`
Statement II If `f(x)` is concave up then `f''(x)gt0`

Options

A. Both statement I and Statement II are correct and Statement II is the correct explanation of Statement I
B. Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I
C. Statement I is correct but Statement II is incorrect
D. Statement II is correct but Statement I is incorrect.

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{x}{x^2 - 1} \) and determine the concavity of its graph for \( x > 1 \). We will also verify the statements provided.

### Step 1: Find the first derivative \( f'(x) \)

Using the quotient rule, where \( u = x \) and \( v = x^2 - 1 \):
\[ f'(x) = \frac{u'v - uv'}{v^2} \]
Here, \( u' = 1 \) and \( v' = 2x \).

Substituting these into the quotient rule:
\[ f'(x) = \frac{(1)(x^2 - 1) - (x)(2x)}{(x^2 - 1)^2} = \frac{x^2 - 1 - 2x^2}{(x^2 - 1)^2} = \frac{-x^2 - 1}{(x^2 - 1)^2} = \frac{-(x^2 + 1)}{(x^2 - 1)^2} \]

### Step 2: Find the second derivative \( f''(x) \)

Now we differentiate \( f'(x) \) using the quotient rule again:
Let \( u = -(x^2 + 1) \) and \( v = (x^2 - 1)^2 \).

Calculating \( u' \) and \( v' \):
\[ u' = -2x, \quad v' = 2(x^2 - 1)(2x) = 4x(x^2 - 1) \]

Using the quotient rule:
\[ f''(x) = \frac{u'v - uv'}{v^2} \]
Substituting the values:
\[ f''(x) = \frac{(-2x)(x^2 - 1)^2 - (-(x^2 + 1))(4x(x^2 - 1))}{(x^2 - 1)^4} \]
Simplifying this expression will give us \( f''(x) \).

### Step 3: Analyze the sign of \( f''(x) \) for \( x > 1 \)

To determine the concavity, we need to check if \( f''(x) > 0 \) for \( x > 1 \).

1. The term \( (x^2 - 1)^2 \) is always positive for \( x > 1 \).
2. We need to analyze the numerator \( -2x(x^2 - 1)^2 + 4x(x^2 + 1)(x^2 - 1) \).

After simplification, we can check if the resulting expression is positive for \( x > 1 \).

### Conclusion

- If \( f''(x) > 0 \) for \( x > 1 \), then the graph of \( f(x) \) is concave up for \( x > 1 \), confirming Statement I.
- Statement II is true as well, since if a function is concave up, then \( f''(x) > 0 \).

Thus, both statements are correct.

### Final Answer

Both Statement I and Statement II are correct.