To find the intervals where the function $g(x) = (x+1)^3$ is increasing or decreasing, we need to follow these steps:

Step 1: Find the derivative of $g(x)$

The derivative $g'(x)$ gives us the slope of the function: $g(x) = (x+1)^3 \quad \implies \quad g'(x) = 3(x+1)^2$

Step 2: Analyze the critical points

Critical points occur where $g'(x) = 0$ or $g'(x)$ is undefined.

• Set $g'(x) = 0$: $3(x+1)^2 = 0 \quad \implies \quad (x+1)^2 = 0 \quad \implies \quad x = -1$

• $g'(x)$ is defined for all $x$, so there are no points where the derivative is undefined.

Step 3: Test intervals around the critical point

The critical point splits the domain into two intervals: $(-\infty, -1)$ and $(-1, \infty)$. We test the sign of $g'(x)$ in each interval.

• For $x \in (-\infty, -1)$: Choose $x = -2$: $g'(-2) = 3((-2)+1)^2 = 3(-1)^2 = 3 > 0$

• For $x \in (-1, \infty)$: Choose $x = 0$: $g'(0) = 3((0)+1)^2 = 3(1)^2 = 3 > 0$

Step 4: Determine behavior

• $g'(x) > 0$ for both intervals $(-\infty, -1)$ and $(-1, \infty)$.
• This means the function is increasing on its entire domain.

Final Answer:

The function $g(x) = (x+1)^3$ is increasing on the open interval $(-\infty, \infty)$ and does not have any intervals where it is decreasing.

Would you like further clarification or a graph of the function?

Related Questions:

1. What are the local extrema of $g(x) = (x+1)^3$, if any?
2. Can you explain how the shape of $g(x)$ relates to the derivative?
3. How does the second derivative help confirm the increasing nature of $g(x)$?
4. How does shifting $g(x) = x^3$ horizontally by 1 unit affect the behavior?
5. What would change if $g(x) = -(x+1)^3$ instead?

Tip:

For polynomial functions, always check for repeated roots in the derivative—they often indicate a plateau rather than a change in direction!