For all `t gt 0 , f(x) = (t^(2)-1)/(t-1)-t` . Which of the following is true about f(t) ?

Options

A. It increases in proportion to t.
B. It increases in proportion to `t^(2)`.
C. It decreases in proportion to `t`.
D. It remains constant .

Answer

To solve the problem, we need to analyze the function given and determine its behavior as \( t \) varies. The function is defined as:

\[ f(t) = \frac{t^2 - 1}{t - 1} - t \]

### Step 1: Simplify the function
We start by simplifying \( f(t) \).

The numerator \( t^2 - 1 \) can be factored using the difference of squares:

\[ t^2 - 1 = (t - 1)(t + 1) \]

Thus, we can rewrite \( f(t) \):

\[ f(t) = \frac{(t - 1)(t + 1)}{t - 1} - t \]

### Step 2: Cancel common terms
Since \( t > 0 \) and \( t \neq 1 \), we can cancel \( t - 1 \) in the numerator and denominator:

\[ f(t) = t + 1 - t \]

### Step 3: Simplify further
Now, we simplify the expression:

\[ f(t) = 1 \]

### Step 4: Analyze the result
The function \( f(t) \) simplifies to a constant value of 1, regardless of the value of \( t \) (as long as \( t > 0 \)).

### Conclusion
Since \( f(t) = 1 \) is constant, we can conclude that:

- \( f(t) \) does not increase or decrease with respect to \( t \).
- Therefore, the correct option is that it remains constant.

### Final Answer
The correct statement about \( f(t) \) is that it remains constant.

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