consider the function `f(x)=(x^(2))/(x^(2)-1)`
f has

Options

A. local maxima but not local minima
B. local minima but not local maxima
C. both local maxima and local minima
D. neither local maxima nor local minima

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{x^2}{x^2 - 1} \) to find its maxima and minima. Here’s a step-by-step solution:

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

To find the critical points where maxima or minima may occur, we first need to differentiate \( f(x) \).

Using the quotient rule, where \( u = x^2 \) and \( v = x^2 - 1 \):

\[ f'(x) = \frac{u'v - uv'}{v^2} \]

Calculating \( u' \) and \( v' \):

- \( u' = 2x \)
- \( v' = 2x \)

Now substituting into the quotient rule:

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

Simplifying the numerator:

\[ f'(x) = \frac{2x(x^2 - 1) - 2x^3}{(x^2 - 1)^2} = \frac{2x(x^2 - 1 - x^2)}{(x^2 - 1)^2} = \frac{-2x}{(x^2 - 1)^2} \]

### Step 2: Set the first derivative equal to zero

To find critical points, set \( f'(x) = 0 \):

\[ \frac{-2x}{(x^2 - 1)^2} = 0 \]

This implies:

\[ -2x = 0 \implies x = 0 \]

### Step 3: Determine the nature of the critical point using the second derivative \( f''(x) \)

Next, we need to find the second derivative \( f''(x) \) to determine whether \( x = 0 \) is a maximum or minimum.

We differentiate \( f'(x) \):

Using the quotient rule again:

\[ f''(x) = \frac{(v^2)(u'') - (u')(v^2)'}{v^4} \]

Where \( u' = -2 \) and \( v = (x^2 - 1)^2 \).

Calculating \( v' \):

\[ v' = 2(x^2 - 1)(2x) = 4x(x^2 - 1) \]

Now substituting back into the formula for \( f''(x) \):

\[ f''(x) = \frac{(x^2 - 1)^2(-2) - (-2)(4x(x^2 - 1))}{(x^2 - 1)^4} \]

Simplifying:

\[ f''(x) = \frac{-2(x^2 - 1)^2 + 8x(x^2 - 1)}{(x^2 - 1)^4} \]

### Step 4: Evaluate \( f''(0) \)

Now we evaluate \( f''(0) \):

\[ f''(0) = \frac{-2(0^2 - 1)^2 + 8(0)(0^2 - 1)}{(0^2 - 1)^4} = \frac{-2(1) + 0}{1} = -2 \]

Since \( f''(0) < 0 \), this indicates that \( x = 0 \) is a local maximum.

### Conclusion

The function \( f(x) = \frac{x^2}{x^2 - 1} \) has a local maximum at \( x = 0 \) and does not have any local minima.