To find the relative extrema of the function $G(x) = \frac{-6}{x^2 + 3}$, we need to follow these steps:

Step 1: Find the derivative

We begin by finding the first derivative of the function $G(x)$ to identify critical points.

$G(x) = \frac{-6}{x^2 + 3}$

We use the quotient rule, which states that if $G(x) = \frac{f(x)}{g(x)}$, then:

$G'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$

Here, $f(x) = -6$ (a constant) and $g(x) = x^2 + 3$. Therefore:

$f'(x) = 0, \quad g'(x) = 2x$

Now apply the quotient rule:

$G'(x) = \frac{(0)(x^2 + 3) - (-6)(2x)}{(x^2 + 3)^2} = \frac{12x}{(x^2 + 3)^2}$

Step 2: Set the derivative equal to zero

To find the critical points, set the first derivative equal to zero and solve for $x$:

$\frac{12x}{(x^2 + 3)^2} = 0$

Since the denominator cannot be zero (because $x^2 + 3 > 0$ for all real $x$), we only need to set the numerator equal to zero:

$12x = 0 \quad \Rightarrow \quad x = 0$

So, $x = 0$ is a critical point.

Step 3: Determine whether it's a minimum or maximum

To classify the critical point $x = 0$ as a relative minimum or maximum, we use the second derivative test or analyze the behavior of $G'(x)$ around $x = 0$.

Second Derivative Test

First, we compute the second derivative $G''(x)$. We start from:

$G'(x) = \frac{12x}{(x^2 + 3)^2}$

Using the quotient rule again, we differentiate this expression to get $G''(x)$. Instead, for simplicity, we analyze the sign of the first derivative $G'(x)$.

• For $x > 0$, $G'(x) > 0$ (positive).
• For $x < 0$, $G'(x) < 0$ (negative).

Thus, $G'(x)$ changes from negative to positive at $x = 0$, indicating that $x = 0$ is a relative minimum.

Step 4: Sketch the graph

The function $G(x) = \frac{-6}{x^2 + 3}$ is a smooth, continuous function for all real values of $x$, and it has a relative minimum at $x = 0$. Since the function approaches zero as $x \to \infty$ or $x \to -\infty$, we know that the graph has the following features:

• Relative minimum at $(0, G(0)) = (0, -2)$.
• No relative maximum.
• The function is negative for all $x$ and approaches 0 from below as $|x| \to \infty$.

Would you like a sketch of this graph?

Summary:

• Relative minimum: at $x = 0$, $G(0) = -2$.
• Relative maximum: None.

Follow-up Questions:

1. What is the quotient rule in calculus and how is it applied?
2. How does the second derivative test determine the nature of critical points?
3. Can you describe why the function has no relative maximum?
4. What are the general characteristics of rational functions like $G(x)$?
5. How does behavior at infinity influence the sketch of $G(x)$?

Tip:

The second derivative test is a quick way to classify extrema without analyzing the entire function behavior.