To find the relative maximum and minimum of the function $g(x)$, we start with the derivative $g'(x) = x^2 + 5x + 4$. The first step is to find the critical points by setting the derivative equal to zero and solving for $x$.

Step 1: Find critical points

Set $g'(x) = 0$:

$x^2 + 5x + 4 = 0$

We can factor the quadratic equation:

$(x + 1)(x + 4) = 0$

Thus, the critical points are:

$x = -1 \quad \text{and} \quad x = -4$

Step 2: Determine the nature of the critical points

To determine whether each critical point is a relative maximum or minimum, we use the second derivative test. First, we need to find the second derivative of $g(x)$.

Step 3: Find the second derivative

Differentiate $g'(x) = x^2 + 5x + 4$:

$g''(x) = 2x + 5$

Step 4: Apply the second derivative test

• For $x = -1$, calculate $g''(-1)$:

$g''(-1) = 2(-1) + 5 = -2 + 5 = 3$

Since $g''(-1) > 0$, this indicates that $x = -1$ is a relative minimum.

• For $x = -4$, calculate $g''(-4)$:

$g''(-4) = 2(-4) + 5 = -8 + 5 = -3$

Since $g''(-4) < 0$, this indicates that $x = -4$ is a relative maximum.

Conclusion

• There is a relative minimum at $x = -1$.
• There is a relative maximum at $x = -4$.

Let me know if you'd like further clarification or a more detailed explanation!

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Related Questions:

1. How do you interpret the second derivative test geometrically?
2. What are some other methods to find relative extrema (besides the second derivative test)?
3. How would you classify the critical points if the second derivative is zero