Question

select the correct answer. this table models continuous function f.

x	-2	-1	0	1	2	3
f(x)	0	-8	-6	0	4	0

if function f is a cubic polynomial, which statement most accurately describes the function over the interval (0, 1)?
a. the function is increasing over the interval (0, 1).
b. the function increases and decreases over the interval (0, 1)
c. the function is decreasing over the interval (0, 1).
d. the function is constant over the interval (0, 1).

Explanation:

Step1: Analyze function values at endpoints

At \( x = 0 \), \( f(0)=-6 \); at \( x = 1 \), \( f(1)=0 \). Also, check values around, like \( x = 0.5 \) (estimating, since it's a cubic, but from table, between \( x=0 \) (f=-6) and \( x=1 \) (f=0), and at \( x=2 \), f=4 (increasing after x=1? Wait, no, interval is (0,1). Wait, at x=0, f=-6; x=1, f=0. But also, check the trend: from x=-2 (f=0) to x=-1 (f=-8) (decreasing), x=-1 to 0 (f=-6) (increasing? Wait no, x=-1: -8, x=0: -6 (so increasing from -1 to 0). Then x=0: -6, x=1: 0 (increasing from 0 to 1? Wait but wait, the interval is (0,1). Wait, but wait, the function is cubic. Wait, but the table: x=0, f=-6; x=1, f=0. But also, at x=2, f=4 (so after x=1, it's increasing). But in (0,1), from x=0 (f=-6) to x=1 (f=0), but wait, is there a decrease in between? Wait, no, the table has x=0,1,2,... Wait, maybe I misread. Wait the table:

x | f(x)
-2 | 0
-1 | -8
0 | -6
1 | 0
2 | 4
3 | 0

Wait, so from x=0 (f=-6) to x=1 (f=0): that's an increase (from -6 to 0). But wait, is there a point where it decreases? Wait, no, the interval is (0,1). Wait, but the function is cubic. Wait, maybe the key is: at x=0, f=-6; x=1, f=0. But also, check the derivative idea (since it's cubic, the slope between 0 and 1: \( \frac{f(1)-f(0)}{1 - 0}=\frac{0 - (-6)}{1}=6>0 \), so increasing? But wait, no, wait the options: option B says "increases and decreases"? Wait no, wait maybe I made a mistake. Wait, wait the table: x=0, f=-6; x=1, f=0. But what about between x=0 and x=1? Wait, the function is continuous (given) and cubic. Wait, but the table has x=0,1,2,... So maybe the function at x=0.5: let's see, the cubic could have a local max or min? Wait, no, from x=0 (f=-6) to x=1 (f=0), and at x=2, f=4 (so increasing after x=1). But in (0,1), is it only increasing? Wait, no, wait the options: option B is "increases and decreases", option A is "increasing". Wait, maybe I misread the table. Wait, x=0: -6, x=1: 0, x=2: 4 (so increasing from 0 to 2). But wait, at x=3, f=0 (so decreasing after x=2). But in (0,1), from x=0 to 1, f goes from -6 to 0, so increasing? But wait, the table at x=0: -6, x=1: 0, x=2:4. So the slope from 0 to 1 is (0 - (-6))/1=6, from 1 to 2 is (4 - 0)/1=4. So both positive, but decreasing slope? Wait, no, the question is about the interval (0,1). Wait, maybe the function has a local minimum or maximum in (0,1)? But the table doesn't have intermediate points. Wait, but the function is cubic, so it can have a local max and min. Wait, but from the table, at x=0, f=-6; x=1, f=0; x=2, f=4. So the function is increasing from x=0 to x=2? But that can't be, because at x=3, f=0 (so it decreases after x=2). Wait, maybe the key is: in (0,1), the function goes from f(0)=-6 to f(1)=0, but is there a point where it decreases? Wait, no, the table doesn't show that. Wait, maybe I made a mistake. Wait, the options: A: increasing, B: increases and decreases, C: decreasing, D: constant.

Wait, let's re-express the function values around (0,1):

At x=0: f=-6

At x=1: f=0

So the average rate of change is positive (0 - (-6))/1 = 6. But since it's a cubic, it could have a local minimum or maximum in (0,1). Wait, but the table doesn't have x=0.5, but let's think about the cubic's behavior. A cubic function has a derivative that's a quadratic, so it can have two critical points (max and min). Let's see the values:

From x=-2 (0) to x=-1 (-8): decreasing

x=-1 (-8) to x=0 (-6): increasing (slope 2)

x=0 (-6) to x=1 (0): slope 6 (increasing)

x=1 (0) to x=2 (4): slope 4 (increasing, but less steep)

x=2 (4) to x=3 (0): slope -4 (decreasing)

So in (0,1), the function is increasing from x=0 to x=1? But wait, the slope from 0 to 1 is 6, and from 1 to 2 is 4, so the slope is decreasing (concave down), but still positive. So the function is increasing on (0,1), but the rate of increase is decreasing (so it's increasing but the slope is decreasing, meaning it's concave down, but still increasing). Wait, but the options: option B says "increases and decreases" – but in (0,1), does it decrease? From the table, f(0)=-6, f(1)=0, so it's increasing. But maybe I'm missing something. Wait, no, the cubic could have a local maximum in (0,1), but the table doesn't show that. Wait, the table only has integer x values. So between x=0 and x=1, the function goes from -6 to 0, so it's increasing. But wait, the answer options: option B is "increases and decreases" – maybe I misread the table. Wait, wait the table: x=0, f=-6; x=1, f=0; x=2, f=4. So from 0 to 1, it's increasing. But maybe the function has a local minimum at some point in (0,1)? No, because f(0)=-6, f(1)=0, so if it had a local minimum, it would be lower than -6, but at x=-1, f=-8 (lower), but (0,1) is between 0 and 1. Wait, maybe the key is: the function is cubic, so it must have a local max and min. But in (0,1), does it have both? Wait, from x=0 (f=-6) to x=1 (f=0), and at x=2 (f=4), so the function is increasing from 0 to 2, but with a decreasing slope (so concave down). But that's still increasing. Wait, but the options: option B is "increases and decreases" – maybe I made a mistake. Wait, no, let's check the options again:

A. The function is increasing over (0,1).

B. The function increases and decreases over (0,1).

C. The function is decreasing over (0,1).

D. The function is constant over (0,1).

From the table, at x=0, f=-6; x=1, f=0 (so increasing). But wait, is there a point where it decreases? For example, suppose at x=0.5, f(x) is less than -6? No, because at x=-1, f=-8 (lower), but (0,1) is after x=0. So between 0 and 1, the function goes from -6 to 0, so it's increasing. But wait, the function is cubic, so its derivative is quadratic. The derivative at x=0: let's estimate, from x=-1 to 0: f(-1)=-8, f(0)=-6, so slope is (-6 - (-8))/1=2. From x=0 to 1: slope is (0 - (-6))/1=6. So the derivative is increasing from x=-1 to 0 (slope 2) to x=0 to 1 (slope 6), meaning the second derivative is positive (concave up) there? Wait, no, slope from -1 to 0 is 2, from 0 to 1 is 6: so derivative is increasing, so second derivative positive (concave up). Then from x=1 to 2: slope 4 (decreasing), so derivative decreasing, second derivative negative (concave down). So in (0,1), the derivative (slope) is increasing from 2 to 6, so the function is increasing, and concave up. But the option B says "increases and decreases" – maybe I misread the table. Wait, wait the table: x=0, f=-6; x=1, f=0; x=2, f=4. So from 0 to 1, it's increasing. But maybe the correct answer is B? Wait, no, maybe I made a mistake. Wait, the function is cubic, so it must have a local maximum and minimum. Let's find the critical points. The function can be modeled. Let's assume the cubic passes through (-2,0), (-1,-8), (0,-6), (1,0), (2,4), (3,0). Let's find the cubic equation. Let \( f(x)=ax^3+bx^2+cx+d \).

Using x=-2: \( -8a + 4b - 2c + d = 0 \)

x=-1: \( -a + b - c + d = -8 \)

x=0: \( d = -6 \)

x=1: \( a + b + c + d = 0 \)

x=2: \( 8a + 4b + 2c + d = 4 \)

x=3: \( 27a + 9b + 3c + d = 0 \)

From x=0, d=-6.

Substitute d=-6 into x=-2: \( -8a + 4b - 2c -6 = 0 \Rightarrow -8a + 4b - 2c = 6 \Rightarrow -4a + 2b - c = 3 \) (1)

x=-1: \( -a + b - c -6 = -8 \Rightarrow -a + b - c = -2 \) (2)

x=1: \( a + b + c -6 = 0 \Rightarrow a + b + c = 6 \) (3)

Subtract (2) from (1): \( (-4a + 2b - c) - (-a + b - c) = 3 - (-2) \Rightarrow -3a + b = 5 \) (4)

Add (2) and (3): \( (-a + b - c) + (a + b + c) = -2 + 6 \Rightarrow 2b = 4 \Rightarrow b=2 \)

From (4): -3a + 2 = 5 \Rightarrow -3a=3 \Rightarrow a=-1

From (3): -1 + 2 + c = 6 \Rightarrow c=5

So the cubic is \( f(x)=-x^3 + 2x^2 + 5x -6 \)

Now, find derivative: \( f'(x)=-3x^2 + 4x + 5 \)

Set derivative to zero: \( -3x^2 + 4x + 5 = 0 \Rightarrow 3x^2 -4x -5 = 0 \)

Solutions: \( x=\frac{4\pm\sqrt{16 + 60}}{6}=\frac{4\pm\sqrt{76}}{6}=\frac{4\pm2\sqrt{19}}{6}=\frac{2\pm\sqrt{19}}{3} \)

\( \sqrt{19}\approx4.358 \), so \( x=\frac{2 + 4.358}{3}\approx2.119 \), \( x=\frac{2 - 4.358}{3}\approx-0.786 \)

So critical points at x≈-0.786 and x≈2.119. So in the interval (0,1), the critical point is at x≈-0.786 (left of 0) and x≈2.119 (right of 1). So in (0,1), the derivative \( f'(x)=-3x^2 + 4x + 5 \). Let's evaluate at x=0: f'(0)=5>0; at x=1: f'(1)=-3 + 4 + 5=6>0. So derivative is positive on (0,1), meaning the function is increasing on (0,1). Wait, but that contradicts option B. But wait, the cubic we found has critical points at x≈-0.786 and x≈2.119, so in (0,1), derivative is positive (since between -0.786 and 2.119, derivative is positive). So the function is increasing on (0,1). But wait, the table at x=0: -6, x=1: 0, x=2:4, which matches the cubic. So why is option B an option? Maybe I made a mistake in the cubic? Wait, let's check x=-2: f(-2)=-(-8) + 2(4) + 5(-2) -6=8 + 8 -10 -6=0, correct. x=-1: -(-1) + 2(1) + 5(-1) -6=1 + 2 -5 -6=-8, correct. x=0: -0 + 0 + 0 -6=-6, correct. x=1: -1 + 2 + 5 -6=0, correct. x=2: -8 + 8 + 10 -6=4, correct. x=3: -27 + 18 + 15 -6=0, correct. So the cubic is correct. Then derivative on (0,1): f'(x)=-3x²+4x+5. At x=0, 5; x=1, 6. So positive, so function is increasing on (0,1). But wait, the options: A is "increasing", B is "increases and decreases". But according to the derivative, it's only increasing. But maybe the question's table was misread? Wait, no, the table is as given. Wait, maybe the answer is B? No, the derivative shows it's increasing. Wait, maybe the problem is that the function is cubic, so it must have a local max and min, but in (0,1), it's between two critical points? No, the critical points are at x≈-0.786 and x≈2.119, so (0,1) is between -0.786 and 2.119, so derivative is positive, so increasing. So the correct answer should be A? But wait, the initial analysis: from x=0 (-6) to x=1 (0), increasing. But maybe the option B is a trick? Wait, no, the derivative is positive throughout (0,1), so it's increasing. So the correct option is A? Wait, but let's check again.

Wait, the cubic function \( f(x)=-x^3 + 2x^2 + 5x -6 \). Let's compute f(0.5): \( -0.125 + 2(0.25) + 5(0.5) -6 = -0.125 + 0.5 + 2.5 -6 = -3.125 \). Wait, that's less than f(0)=-6? No, -3.125 is greater than -6. Wait, f(0)=-6, f(0.5)=-3.125 (increasing), f(1)=0 (increasing). So yes, increasing on (0,1). So the correct answer is A? But wait, the options: A is "The function is increasing over the interval (0, 1)".

Step2: Confirm with derivative and function values

Since the derivative \( f'(x) \) is positive on (0,1) (as shown by solving the cubic's derivative and evaluating at endpoints of (0,1)), the function is increasing on (0,1).

Answer:

A. The function is increasing over the interval (0, 1)