Question

use the graph of f(x) below to put the values of f(-6), f(0), and f(4) in decreasing order.options: f(-6) > f(0) > f(4); f(-6) > f(4) > f(0); f(0) > f(4) > f(-6); f(0) > f(-6) > f(4); f(4) > f(-6) > f(0); f(4) > f(0) > f(-6)

Explanation:

Step1: Recall the meaning of the derivative

The derivative \( f'(x) \) at a point \( x \) represents the slope of the tangent line to the graph of \( f(x) \) at that point. A steeper positive slope means a larger value of \( f'(x) \), and a flatter slope (closer to zero) means a smaller value of \( f'(x) \).

Step2: Analyze the slope at \( x = -6 \)

At \( x = -6 \), the graph of \( f(x) \) is very flat (almost horizontal). So the slope of the tangent line here, \( f'(-6) \), is close to 0 (a small positive number, since the function is increasing very slowly).

Step3: Analyze the slope at \( x = 0 \)

At \( x = 0 \), the graph is steeper than at \( x = -6 \). The curve is increasing more rapidly here, so the slope \( f'(0) \) is larger than \( f'(-6) \).

Step4: Analyze the slope at \( x = 4 \)

At \( x = 4 \), the graph is flatter than at \( x = 0 \) (it's approaching a horizontal asymptote, maybe? The curve is increasing but very slowly, almost flat). So the slope \( f'(4) \) is smaller than \( f'(0) \) and, since the curve is still increasing (just slowly), it's positive but smaller than \( f'(0) \) and maybe even smaller than \( f'(-6) \)? Wait, no—wait, let's re-examine. Wait, at \( x = -6 \), the function is almost flat (slope near 0). At \( x = 0 \), the slope is steeper (larger positive). At \( x = 4 \), the slope is flatter than at \( x = 0 \), but is it flatter than at \( x = -6 \)? Wait, the graph at \( x = -6 \) is very flat (almost horizontal), at \( x = 0 \) it's steeper, and at \( x = 4 \) it's still increasing but approaching a horizontal line, so the slope at \( x = 4 \) is smaller than at \( x = 0 \), and the slope at \( x = -6 \) is also small (close to 0), but let's compare the slopes. Wait, the key is: the derivative \( f'(x) \) is the slope of the tangent. So when the graph is increasing, \( f'(x) > 0 \). The steeper the increase, the larger \( f'(x) \). So:

- At \( x = -6 \): very flat, so \( f'(-6) \) is a small positive number (close to 0).
- At \( x = 0 \): steeper, so \( f'(0) \) is a larger positive number.
- At \( x = 4 \): flatter than at \( x = 0 \), so \( f'(4) \) is a positive number but smaller than \( f'(0) \). Now, is \( f'(4) \) smaller than \( f'(-6) \)? Wait, at \( x = -6 \), the graph is almost horizontal (slope ~0), and at \( x = 4 \), the graph is still increasing but very slowly (slope ~0 but maybe slightly larger than at \( x = -6 \))? Wait, no—wait, the graph at \( x = -6 \) is a horizontal line (almost), so slope 0. At \( x = 0 \), slope is positive and steeper. At \( x = 4 \), slope is positive but less steep than at \( x = 0 \), but is it more or less steep than at \( x = -6 \)? Wait, the graph at \( x = -6 \) is horizontal (slope 0), at \( x = 4 \), the graph is still increasing (so slope positive) but very slowly. So \( f'(-6) \) is ~0, \( f'(0) \) is larger, \( f'(4) \) is positive but smaller than \( f'(0) \) and maybe smaller than \( f'(-6) \)? Wait, no—if at \( x = -6 \) the slope is 0 (or almost 0), and at \( x = 4 \) the slope is still positive (since the function is increasing), then \( f'(4) \) is positive, \( f'(-6) \) is ~0, so \( f'(4) > f'(-6) \)? Wait, that contradicts my earlier thought. Wait, let's look at the graph again. The graph starts at the left (x=-6) as a horizontal line (slope 0), then curves up, getting steeper until around x=0, then starts to flatten out as x increases (approaching y=2). So the slope (derivative) starts at 0 (x=-6), increases to a maximum at some point (maybe around x=0), then decreases back towards 0 as x increases (since the curve flattens). So:

- \( f'(-6) \): slope ~0 (small positive)
- \( f'(0) \): slope is larger (steeper)
- \( f'(4) \): slope is smaller than at x=0, but since the curve is still increasing, it's positive, but is it larger or smaller than \( f'(-6) \)? Wait, if the slope increases from x=-6 to x=0 (reaching a peak), then decreases from x=0 to x=4 (towards 0), then:

\( f'(-6) < f'(0) \), and \( f'(4) < f'(0) \), and since the slope decreases after x=0, \( f'(4) \) is less than \( f'(0) \), but is \( f'(4) \) greater than or less than \( f'(-6) \)?

Wait, at x=-6, slope is 0 (or almost 0). At x=4, slope is still positive (since the function is increasing), so \( f'(4) > 0 \), and \( f'(-6) \) is ~0, so \( f'(4) > f'(-6) \)? But that can't be, because the graph at x=4 is flatter than at x=0, but is it flatter than at x=-6? Wait, no—at x=-6, the graph is horizontal (slope 0), at x=4, the graph is still increasing (slope positive), so \( f'(4) \) is positive, \( f'(-6) \) is ~0, so \( f'(4) > f'(-6) \). But then the order would be \( f'(0) > f'(4) > f'(-6) \)? Wait, but let's check the options.

Wait the options are:

1. \( f'(-6) > f'(0) > f'(4) \)
2. \( f'(-6) > f'(4) > f'(0) \)
3. \( f'(0) > f'(4) > f'(-6) \)
4. \( f'(0) > f'(-6) > f'(4) \)
5. \( f'(4) > f'(-6) > f'(0) \)
6. \( f'(4) > f'(0) > f'(-6) \)

Wait, let's re-express the slope behavior. The derivative \( f'(x) \) is the slope of the tangent. The graph is a curve that starts flat (x=-6), becomes steeper (increasing slope) until some point (maybe x=0), then becomes less steep (decreasing slope) as x increases (approaching horizontal). So the slope (f'(x)) increases from x=-6 to x=0, then decreases from x=0 to x=4. So:

- At x=-6: slope = m1 (small, ~0)
- At x=0: slope = m2 (m2 > m1)
- At x=4: slope = m3 (m3 < m2, and since the slope is decreasing after x=0, m3 could be less than m1 or greater? Wait, if the slope increases from m1 (x=-6) to m2 (x=0), then decreases to m3 (x=4), then m3 could be greater than m1 (if the decrease is not too much) or less. But looking at the graph, at x=4, the curve is almost flat (like approaching a horizontal line), so m3 is close to 0, maybe even less than m1? Wait, no—if at x=-6 the slope is 0, and at x=4 the slope is still positive (since the function is increasing), then m3 is positive, so m3 > m1 (since m1 is ~0). Wait, but the graph at x=-6 is horizontal (slope 0), at x=4 it's still increasing (slope positive), so m3 > 0 = m1. Then the order would be m2 (x=0) > m3 (x=4) > m1 (x=-6), which is \( f'(0) > f'(4) > f'(-6) \), which is option 3.

Wait, that makes sense. Because the slope increases from x=-6 (slope 0) to x=0 (steeper slope), then decreases from x=0 to x=4 (slope becomes less steep, but still positive, so slope at x=4 is positive but less than at x=0, and since at x=-6 slope is 0, slope at x=4 is greater than 0, so \( f'(4) > f'(-6) \). Therefore, the order is \( f'(0) > f'(4) > f'(-6) \).

Answer:

\( f'(0) > f'(4) > f'(-6) \) (which corresponds to the option: \( f'(0) > f'(4) > f'(-6) \))