Question

comparación de tasas de cambio
en el gráfico, ignorando si está aumentando o disminuyendo, ¿qué intervalo tiene la mayor tasa de cambio promedio?
x = 2 y x = 4
x = 2 y x = 6
x = 1 y x = 5

Explanation:

To determine the interval with the greatest average rate of change, we use the formula for the average rate of change: $\frac{f(b) - f(a)}{b - a}$, where $a$ and $b$ are the endpoints of the interval.

Step 1: Analyze the graph to find the coordinates of the points at the given x-values.

- For the interval $x = 2$ to $x = 4$:
- Let's assume from the graph that at $x = 2$, $y = 10$ (since the graph reaches a peak around there) and at $x = 4$, $y = 10$ (assuming symmetry or the graph's shape). Then the average rate of change is $\frac{10 - 10}{4 - 2} = \frac{0}{2} = 0$.
- For the interval $x = 2$ to $x = 6$:
- At $x = 2$, $y = 10$; at $x = 6$, $y = 0$ (since the graph touches the x-axis at $x = 6$). The average rate of change is $\frac{0 - 10}{6 - 2} = \frac{-10}{4} = -2.5$.
- For the interval $x = 1$ to $x = 5$:
- At $x = 1$, let's say $y = 8$ (from the graph's grid) and at $x = 5$, $y = 8$ (assuming symmetry). Wait, no, maybe better to check the actual change. Wait, maybe I made a mistake. Wait, the graph starts at (0,0), goes up, peaks, then comes down to (6,0) maybe? Wait, the left endpoint is at (0,0) and the right at (6,0) maybe? Let's re-examine.
Wait, the graph is a parabola opening downward. Let's find the coordinates:
- At $x = 1$: Let's see the y-value. The grid has y-axis with 2,4,6,8,10,12. At $x = 1$, the point is at y = 8? Wait, no, at $x = 0$, it's (0,0). At $x = 1$, maybe (1, 8)? At $x = 2$, (2, 10)? At $x = 3$, peak (3, 11)? Then at $x = 4$, (4, 10), $x = 5$, (5, 8), $x = 6$, (6, 0). Wait, maybe. Let's recalculate:
- Interval $x=2$ to $x=4$: $f(2)=10$, $f(4)=10$. So $\frac{10 - 10}{4 - 2} = 0$.
- Interval $x=2$ to $x=6$: $f(2)=10$, $f(6)=0$. So $\frac{0 - 10}{6 - 2} = -2.5$.
- Interval $x=1$ to $x=5$: $f(1)=8$, $f(5)=8$. Wait, no, that can't be. Wait, maybe $x=1$: (1, 8), $x=5$: (5, 8). Then $\frac{8 - 8}{5 - 1} = 0$. Wait, that's not right. Wait, maybe the left endpoint is (0,0) and right is (6,0). Let's check the slope between (0,0) and (6,0): 0. Between (0,0) and (3,11): slope is 11/3 ≈ 3.67. Between (3,11) and (6,0): slope is -11/3 ≈ -3.67. Wait, maybe the intervals are different. Wait, the question is about which interval has the greatest average rate of change (ignoring if it's increasing or decreasing, so we take the absolute value? Wait, the problem says "ignorando si está aumentando o disminuyendo", so we consider the magnitude? Wait, no, the average rate of change is $\frac{f(b)-f(a)}{b - a}$, regardless of sign. Wait, but maybe the graph is a parabola with vertex at (3, 11), passing through (0,0) and (6,0). Let's confirm the coordinates:
- At $x = 0$, $y = 0$ (left endpoint).
- At $x = 1$, $y = 8$ (since from (0,0) to (3,11), the slope is 11/3 ≈ 3.67, so at x=1, y= 3.67*1 ≈ 3.67? Wait, no, maybe the grid is such that each square is 1 unit. Let's look at the graph: the left red dot is at (0,0), the right at (6,0). The curve goes up to a peak, maybe at (3, 11). Then at x=2, y=10; x=4, y=10; x=1, y=8; x=5, y=8.

Now, let's calculate the average rate of change for each interval:

1. Interval $x = 2$ to $x = 4$:
- $f(2) = 10$, $f(4) = 10$
- Average rate of change: $\frac{10 - 10}{4 - 2} = \frac{0}{2} = 0$

2. Interval $x = 2$ to $x = 6$:
- $f(2) = 10$, $f(6) = 0$
- Average rate of change: $\frac{0 - 10}{6 - 2} = \frac{-10}{4} = -2.5$

3. Interval $x = 1$ to $x = 5$:
- $f(1) = 8$, $f(5) = 8$
- Average rate of change: $\frac{8 - 8}{5 - 1} = \frac{0}{4} = 0$

Wait, that can't be right. Maybe I misread the intervals. Wait, the options are:

- $x = 2$ y $x = 4$
- $x = 2$ y $x = 6$
- $x = 1$ y $x = 5$

Wait, maybe the graph is different. Let's re-express. Maybe the left endpoint is (0,0), and at x=1, y=4; x=2, y=8; x=3, y=10; x=4, y=8; x=5, y=4; x=6, y=0. Ah, that makes more sense. So a parabola symmetric about x=3.

So:

- At x=0: (0,0)
- x=1: (1,4)
- x=2: (2,8)
- x=3: (3,10)
- x=4: (4,8)
- x=5: (5,4)
- x=6: (6,0)

Now let's recalculate:

1. Interval $x = 2$ to $x = 4$:
- $f(2) = 8$, $f(4) = 8$
- Average rate of change: $\frac{8 - 8}{4 - 2} = \frac{0}{2} = 0$

2. Interval $x = 2$ to $x = 6$:
- $f(2) = 8$, $f(6) = 0$
- Average rate of change: $\frac{0 - 8}{6 - 2} = \frac{-8}{4} = -2$

3. Interval $x = 1$ to $x = 5$:
- $f(1) = 4$, $f(5) = 4$
- Average rate of change: $\frac{4 - 4}{5 - 1} = \frac{0}{4} = 0$

Wait, that's still not right. Maybe the peak is higher. Wait, the y-axis has 12 at the top. Maybe at x=3, y=12. Then:

- x=0: (0,0)
- x=1: (1, 6)
- x=2: (2, 10)
- x=3: (3, 12)
- x=4: (4, 10)
- x=5: (5, 6)
- x=6: (6, 0)

Now calculate:

1. Interval $x = 2$ to $x = 4$:
- $f(2) = 10$, $f(4) = 10$
- Average rate of change: $\frac{10 - 10}{4 - 2} = 0$

2. Interval $x = 2$ to $x = 6$:
- $f(2) = 10$, $f(6) = 0$
- Average rate of change: $\frac{0 - 10}{6 - 2} = -2.5$

3. Interval $x = 1$ to $x = 5$:
- $f(1) = 6$, $f(5) = 6$
- Average rate of change: $\frac{6 - 6}{5 - 1} = 0$

Wait, this is confusing. Maybe the question is about the interval with the greatest magnitude of the average rate of change. Let's check the absolute values:

- For $x=2$ to $x=4$: |0| = 0
- For $x=2$ to $x=6$: |-2.5| = 2.5
- For $x=1$ to $x=5$: |0| = 0

So the interval $x = 2$ y $x = 6$ has the greatest average rate of change (in magnitude, since we ignore if it's increasing or decreasing).

Answer:

$x = 2$ y $x = 6$