To find the absolute extrema of the function $g(x) = \frac{x - 3}{x^2}$ on the interval $[-2, 2]$, we need to follow these steps:

1. Find the first derivative of $g(x)$:

The function $g(x) = \frac{x - 3}{x^2}$ is a rational function, so we'll apply the quotient rule to differentiate it. The quotient rule states:

$\frac{d}{dx} \left( \frac{f(x)}{h(x)} \right) = \frac{h(x)f'(x) - f(x)h'(x)}{[h(x)]^2}$

Let $f(x) = x - 3$ and $h(x) = x^2$.

• The derivative of $f(x)$ is $f'(x) = 1$.
• The derivative of $h(x)$ is $h'(x) = 2x$.

Using the quotient rule:

$g'(x) = \frac{x^2(1) - (x - 3)(2x)}{x^4} = \frac{x^2 - 2x(x - 3)}{x^4}$

Simplifying the numerator:

$g'(x) = \frac{x^2 - 2x^2 + 6x}{x^4} = \frac{-x^2 + 6x}{x^4}$

So the first derivative is:

$g'(x) = \frac{-x^2 + 6x}{x^4}$

2. Find the critical points:

Critical points occur where $g'(x) = 0$ or where $g'(x)$ does not exist.

First, set $g'(x) = 0$:

$\frac{-x^2 + 6x}{x^4} = 0$

For the fraction to be zero, the numerator must be zero:

$-x^2 + 6x = 0$

Factoring the equation:

$x(-x + 6) = 0$

So the critical points are:

$x = 0 \quad \text{or} \quad x = 6$

Since $x = 6$ is not within the interval $[-2, 2]$, we discard it. Therefore, the only critical point within the interval is $x = 0$.

3. Check the endpoints of the interval:

Next, evaluate the function at the endpoints of the interval, $x = -2$ and $x = 2$.

• At $x = -2$: $g(-2) = \frac{-2 - 3}{(-2)^2} = \frac{-5}{4}$

• At $x = 2$: $g(2) = \frac{2 - 3}{(2)^2} = \frac{-1}{4}$

4. Evaluate the function at the critical point $x = 0$:

• At $x = 0$: $g(0) = \frac{0 - 3}{0^2}$ This is undefined due to division by zero.

5. Determine the absolute extrema:

Since the function is undefined at $x = 0$, we do not consider it a valid point for extrema. We now compare the function values at the endpoints of the interval:

• $g(-2) = \frac{-5}{4} = -1.25$
• $g(2) = \frac{-1}{4} = -0.25$

Thus, the absolute maximum value is $-0.25$ at $x = 2$, and the absolute minimum value is $-1.25$ at $x = -2$.

Final Answer:

• Absolute maximum: $g(2) = -0.25$ at $x = 2$
• Absolute minimum: $g(-2) = -1.25$ at $x = -2$

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Would you like further clarification or more detailed steps on any part of the process?

Related Questions:

1. How do you use the second derivative test to classify critical points?
2. What is the difference between absolute extrema and local extrema?
3. How can you apply the first derivative test to determine intervals of increase and decrease for a function?
4. How do you handle functions with undefined points within a given interval?
5. How do you find critical points for higher-order rational functions?

Tip: When dealing with rational functions, always check for points where the denominator equals zero, as these will be points where the function is undefined and could lead to vertical asymptotes or critical behavior.