The function `f(x)=|x^2-2|/|x^2-4|` has, (a) no point of local maxima (b) no point of local minima (c) exactly one point of local minima (d) exactly one point of local maxima

Options

A. no point of local minima
B. no point of local minima
C. exactly one point of local minima
D. exaclty one point of local maxima

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{|x^2 - 2|}{|x^2 - 4|} \) to determine the points of local maxima and minima.

### Step 1: Identify the points where the function is undefined
The function \( f(x) \) is undefined where the denominator is zero. Thus, we need to find the values of \( x \) for which \( |x^2 - 4| = 0 \).

\[ x^2 - 4 = 0 \implies x^2 = 4 \implies x = 2 \text{ or } x = -2 \]

**Hint:** Check where the denominator becomes zero to identify points of discontinuity.

### Step 2: Analyze the function
The function can be rewritten as:

\[ f(x) = \frac{|x^2 - 2|}{|x^2 - 4|} = \frac{|(x - \sqrt{2})(x + \sqrt{2})|}{|(x - 2)(x + 2)|} \]

This shows that the function has critical points at \( x = \sqrt{2}, -\sqrt{2}, 2, -2 \). However, \( x = 2 \) and \( x = -2 \) are not in the domain of \( f(x) \).

**Hint:** Rewrite the function in factored form to identify critical points.

### Step 3: Differentiate the function
To find local maxima and minima, we need to differentiate \( f(x) \) using the quotient rule. The derivative is given by:

\[ f'(x) = \frac{(g(x) \cdot h'(x) - h(x) \cdot g'(x))}{(h(x))^2} \]

where \( g(x) = |x^2 - 2| \) and \( h(x) = |x^2 - 4| \).

### Step 4: Find critical points
Set \( f'(x) = 0 \) to find critical points. The critical points occur when the numerator of the derivative is zero.

\[ g(x) \cdot h'(x) - h(x) \cdot g'(x) = 0 \]

From our earlier analysis, we can find that the only root of this equation is \( x = 0 \).

**Hint:** Solve \( f'(x) = 0 \) to find critical points.

### Step 5: Determine the nature of critical points
To determine whether \( x = 0 \) is a local maxima or minima, we can analyze the sign of \( f'(x) \) around this point.

- For \( x < 0 \): \( f'(x) > 0 \) (function is increasing)
- For \( x > 0 \): \( f'(x) < 0 \) (function is decreasing)

Since the function changes from increasing to decreasing at \( x = 0 \), we conclude that \( x = 0 \) is a local maximum.

**Hint:** Use the first derivative test to determine the nature of the critical points.

### Conclusion
The function \( f(x) = \frac{|x^2 - 2|}{|x^2 - 4|} \) has exactly one point of local maxima at \( x = 0 \).

Thus, the correct answer is:
(d) exactly one point of local maxima.