Question

f(x) = x³ - x - 1
over which interval does f have an average rate of change of zero?
choose 1 answer:
a -1 ≤ x ≤ 2
b 2 ≤ x ≤ 3
c -5 ≤ x ≤ 5
d -3 ≤ x ≤ -2
show calculator

Explanation:

The average rate of change of a function \( f(x) \) over the interval \([a, b]\) is given by the formula:
\[ \text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a} \]
We need to find the interval \([a, b]\) where this average rate of change is zero. This means we need \( f(b) = f(a) \) (since if \( \frac{f(b) - f(a)}{b - a} = 0 \), then \( f(b) - f(a) = 0 \) as long as \( b eq a \)).

Let's analyze each option:

Option A: \(-1 \leq x \leq 2\)

First, calculate \( f(-1) \):
\[ f(-1) = (-1)^3 - (-1) - 1 = -1 + 1 - 1 = -1 \]
Then, calculate \( f(2) \):
\[ f(2) = (2)^3 - 2 - 1 = 8 - 2 - 1 = 5 \]
Since \( f(-1) eq f(2) \), the average rate of change is not zero.

Option B: \(2 \leq x \leq 3\)

Calculate \( f(2) \):
\[ f(2) = 8 - 2 - 1 = 5 \]
Calculate \( f(3) \):
\[ f(3) = 27 - 3 - 1 = 23 \]
Since \( f(2) eq f(3) \), the average rate of change is not zero.

Option C: \(-5 \leq x \leq 5\)

Calculate \( f(-5) \):
\[ f(-5) = (-5)^3 - (-5) - 1 = -125 + 5 - 1 = -121 \]
Calculate \( f(5) \):
\[ f(5) = 5^3 - 5 - 1 = 125 - 5 - 1 = 119 \]
Since \( f(-5) eq f(5) \), the average rate of change is not zero.

Option D: \(-3 \leq x \leq -2\)

Calculate \( f(-3) \):
\[ f(-3) = (-3)^3 - (-3) - 1 = -27 + 3 - 1 = -25 \]
Calculate \( f(-2) \):
\[ f(-2) = (-2)^3 - (-2) - 1 = -8 + 2 - 1 = -7 \]
Wait, that can't be right. Wait, maybe I made a mistake. Wait, no, let's re - calculate.

Wait, the function is \( f(x)=x^{3}-x - 1 \)

For \( x=-3 \):
\( f(-3)=(-3)^{3}-(-3)-1=-27 + 3-1=-25 \)

For \( x = - 2 \):
\( f(-2)=(-2)^{3}-(-2)-1=-8 + 2-1=-7 \)

Wait, that's not equal. Wait, maybe I misread the options. Wait, the original function is \( f(x)=x^{3}-x - 1 \). Wait, maybe there is a mistake in my calculation. Wait, no, let's check again.

Wait, the average rate of change is \( \frac{f(b)-f(a)}{b - a} \). We need \( f(b)-f(a)=0 \), so \( f(b)=f(a) \)

Let's check the function \( f(x)=x^{3}-x - 1 \). Let's see if it's odd or even. A function is odd if \( f(-x)=-f(x) \), even if \( f(-x)=f(x) \)

\( f(-x)=(-x)^{3}-(-x)-1=-x^{3}+x - 1 \)

\( -f(x)=-x^{3}+x + 1 \)

So it's neither odd nor even. Wait, maybe I made a mistake in the options. Wait, the option C is \(-5\leq x\leq5\), let's recalculate \( f(-5) \) and \( f(5) \)

\( f(-5)=(-5)^{3}-(-5)-1=-125 + 5-1=-121 \)

\( f(5)=5^{3}-5 - 1=125-5 - 1=119 \). Not equal.

Wait, option D: \( - 3\leq x\leq - 2 \)

Wait, maybe I made a mistake in the function. Wait, the function is \( f(x)=x^{3}-x - 1 \). Let's check the derivative to see where it's increasing or decreasing. The derivative \( f^\prime(x)=3x^{2}-1 \). The critical points are at \( x=\pm\frac{1}{\sqrt{3}}\approx\pm0.577 \)

So the function is increasing when \( |x|>\frac{1}{\sqrt{3}} \) and decreasing when \( |x|<\frac{1}{\sqrt{3}} \)

Now, let's check the values again.

Wait, maybe the question is mis - printed? Or maybe I made a mistake. Wait, let's check option D again.

\( f(-3)=(-3)^{3}-(-3)-1=-27 + 3-1=-25 \)

\( f(-2)=(-2)^{3}-(-2)-1=-8 + 2-1=-7 \)

\( \frac{f(-2)-f(-3)}{-2-(-3)}=\frac{-7-(-25)}{1}=\frac{18}{1}=18 eq0 \)

Option A: \( f(-1)=(-1)^{3}-(-1)-1=-1 + 1-1=-1 \)

\( f(2)=8 - 2-1=5 \)

\( \frac{5-(-1)}{2-(-1)}=\frac{6}{3}=2 eq0 \)

Option B: \( f(2)=5 \), \( f(3)=27 - 3-1=23 \)

\( \frac{23 - 5}{3 - 2}=18 eq0 \)

Option C: \( f(-5)=-121 \), \( f(5)=119 \)

\( \frac{119-(-121)}{5-(-5)}=\frac{240}{10}=24 eq0 \)

Wait, this is strange. Maybe there is a mistake in the problem or in my calculation. Wait, maybe the function is \( f(x)=x^{3}-x + 1 \)? Let's check. If \( f(x)=x^{3}-x + 1 \), then \( f(-x)=-x^{3}+x + 1=- (x^{3}-x - 1)=-f(x)+2 \), still not odd or even.

Wait, maybe the original function is \( f(x)=x^{3}-x \). Let's check. If \( f(x)=x^{3}-x \), then \( f(-x)=-x^{3}+x=-f(x) \), so it's an odd function. Then \( f(-a)=-f(a) \), so \( f(b)-f(a)=0 \) when \( b=-a \). So for interval \([-5,5]\), \( f(5)-f(-5)=f(5)+f(5)=2f(5) eq0 \) (unless \( f(5)=0 \), but \( f(5)=125 - 5 = 120 eq0 \)).

Wait, maybe the question is about the average rate of change being zero, which requires \( f(b)=f(a) \). Let's check the function \( f(x)=x^{3}-x - 1 \) again.

Wait, maybe I made a mistake in option D. Wait, \( x=-3 \) and \( x=-2 \):

\( f(-3)=(-3)^3-(-3)-1=-27 + 3-1=-25 \)

\( f(-2)=(-2)^3-(-2)-1=-8 + 2-1=-7 \)

No, not equal.

Wait, maybe the option C is \(-1\leq x\leq1\)? But in the given options, it's \(-5\leq x\leq5\).

Wait, maybe there is a typo in the problem. But according to the given options, none of them seem to have \( f(b)=f(a) \). But since this is a multiple - choice question, maybe I made a mistake.

Wait, let's re - calculate the average rate of change for each option:

**Option A: \(-1\leq x\leq2\)**

\( f(-1)=(-1)^3-(-1)-1=-1 + 1-1=-1 \)

\( f(2)=2^3-2 - 1=8 - 2-1=5 \)

Average rate of change \(=\frac{5-(-1)}{2-(-1)}=\frac{6}{3}=2 eq0 \)

**Option B: \(2\leq x\leq3\)**

\( f(2)=5 \), \( f(3)=3^3-3 - 1=27 - 3-1=23 \)

Average rate of change \(=\frac{23 - 5}{3 - 2}=18 eq0 \)

**Option C: \(-5\leq x\leq5\)**

\( f(-5)=(-5)^3-(-5)-1=-125 + 5-1=-121 \)

\( f(5)=5^3-5 - 1=125-5 - 1=119 \)

Average rate of change \(=\frac{119-(-121)}{5-(-5)}=\frac{240}{10}=24 eq0 \)

**Option D: \(-3\leq x\leq - 2\)**

\( f(-3)=(-3)^3-(-3)-1=-27 + 3-1=-25 \)

\( f(-2)=(-2)^3-(-2)-1=-8 + 2-1=-7 \)

Average rate of change \(=\frac{-7-(-25)}{-2-(-3)}=\frac{18}{1}=18 eq0 \)

This is very strange. Maybe the function is \( f(x)=x^{2}-x - 1 \)? Let's check. If \( f(x)=x^{2}-x - 1 \), then it's a quadratic function. The average rate of change is zero when \( f(a)=f(b) \), i.e., when \( a \) and \( b \) are symmetric about the vertex. The vertex of \( y = x^{2}-x - 1 \) is at \( x=\frac{1}{2} \).

For option A: \( x=-1 \) and \( x = 2 \). The mid - point of \(-1\) and \(2\) is \( \frac{-1 + 2}{2}=\frac{1}{2} \), which is the vertex. So \( f(-1)=(-1)^2-(-1)-1=1 + 1-1=1 \), \( f(2)=2^2-2 - 1=4 - 2-1=1 \). So \( f(-1)=f(2) \), and the average rate of change is \( \frac{1 - 1}{2-(-1)} = 0 \).

Ah! Maybe there is a typo in the function, and it's \( f(x)=x^{2}-x - 1 \) instead of \( f(x)=x^{3}-x - 1 \). Assuming that the function is \( f(x)=x^{2}-x - 1 \) (a quadratic function), then for the interval \(-1\leq x\leq2\), the average rate of change is zero.

Answer:

A. \(-1 \leq x \leq 2\)