AP Precalc: Increasing/Decreasing Intervals in Figure 1.15

Answer

Based on the visual evidence in Figure 1.15, the function exhibits a "W-shaped" behavior characteristic of a quartic polynomial. It is decreasing where the graph falls from left to right, specifically on $(-\infty, -1) \cup (0, 2)$, and increasing where the graph rises, on $(-1, 0) \cup (2, \infty)$.

Explanation

I observe a continuous, smooth curve plotted on a Cartesian plane in Figure 1.15. The graph has three local extrema (turning points) located at $x = -1$ (minimum), $x = 0$ (maximum), and $x = 2$ (minimum). The function's tails both point upward toward positive infinity, suggesting an even-degree polynomial with a positive leading coefficient.

1. Identify Intervals of Decrease As we move from left to right along the x-axis, the $y$-values decrease whenever the slope of the tangent line is negative. Looking at Figure 1.15, the graph falls from $x = -\infty$ until it reaches the first local minimum at $x = -1$, and falls again after the local maximum at $x = 0$ until the second local minimum at $x = 2$. The condition for a decreasing interval is: $f'(x) < 0$ This mathematical inequality states that the instantaneous rate of change (the derivative) is negative.

2. Identify Intervals of Increase The function increases whenever the graph moves upward from left to right. This occurs between the first local minimum and the local maximum, and again after the final local minimum. From the grid, these intervals are identified as starting at $x = -1$ and ending at $x = 0$, then starting again at $x = 2$ and continuing to infinity. The condition for an increasing interval is: $f'(x) > 0$ This inequality indicates that the function's output increases as its input increases. ⚠️ This step is required on exams: You must use parentheses (open intervals) for increasing/decreasing behavior because the rate of change is zero at the exact turning point.

3. Relate Behavior to Concavity Concepts Although the text below the image introduces concavity, we can observe it in Figure 1.15. The graph is "concave up" (shaped like a cup) near the minima and "concave down" (shaped like a frown) near the local maximum at the origin. The relationship between concavity and the rate of change is: $\frac{d^2y}{dx^2} > 0 \implies \text{Concave Up}$ A positive second derivative means the slope of the function is increasing over that interval.

Behavior	Graphical Appearance	Mathematical Condition
Increasing	Rising left-to-right	$f'(x) > 0$
Decreasing	Falling left-to-right	$f'(x) < 0$
Concave Up	"Cup" shape	$f''(x) > 0$
Concave Down	"Frown" shape	$f''(x) < 0$

Final Answer

The intervals of increase and decrease for the function in Figure 1.15 are: $\text{Decreasing: } (-\infty, -1) \cup (0, 2)$ $\text{Increasing: } (-1, 0) \cup (2, \infty)$

Common Mistakes

• Using y-values for intervals: Students often mistakenly write the range of $y$-values (e.g., $( -2.5, 0)$) when asked for intervals. Always use $x$-coordinates to describe where a behavior occurs.
• Including endpoints: Using square brackets $[a, b]$ for increasing/decreasing intervals is often penalized in AP Precalculus, as the function is neither increasing nor decreasing at the stationary point where the slope is exactly zero.