Question

1. fill in the table of values.

$f(x) < \sqrt3{(x - 1)}$
$x$ $f(x) = \sqrt3{(x - 1)}$
0
1
9
2
-7
a. $-2$
b. $2$
c. $1$
d. $3$
e. $-3$
f. $0$
g. $-1$
h. $4$

1. choose the best answer.

here is the graph of $f(x) = x^{\frac{1}{3}}$. which graph shows $g(x) = (x - 2)^{\frac{1}{3}}$

Explanation:

Step1: For \( x = 0 \)

Substitute \( x = 0 \) into \( f(x)=\sqrt[3]{x - 1} \). So \( f(0)=\sqrt[3]{0 - 1}=\sqrt[3]{-1}=-1 \). But looking at options, wait, maybe I miscalculated? Wait, no, let's check again. Wait, the options have -1 (option g). Wait, but let's do each \( x \):

Step2: For \( x = 1 \)

Substitute \( x = 1 \) into \( f(x)=\sqrt[3]{1 - 1}=\sqrt[3]{0}=0 \) (option f).

Step3: For \( x = 9 \)

Substitute \( x = 9 \): \( f(9)=\sqrt[3]{9 - 1}=\sqrt[3]{8}=2 \) (option b).

Step4: For \( x = 2 \)

Substitute \( x = 2 \): \( f(2)=\sqrt[3]{2 - 1}=\sqrt[3]{1}=1 \) (option c).

Step5: For \( x = -7 \)

Substitute \( x = -7 \): \( f(-7)=\sqrt[3]{-7 - 1}=\sqrt[3]{-8}=-2 \) (option a).

Wait, let's re - check each:

• \( x = 0 \): \( f(0)=\sqrt[3]{0 - 1}=\sqrt[3]{-1}=-1 \) (option g)
• \( x = 1 \): \( f(1)=\sqrt[3]{1 - 1}=0 \) (option f)
• \( x = 9 \): \( f(9)=\sqrt[3]{9 - 1}=\sqrt[3]{8}=2 \) (option b)
• \( x = 2 \): \( f(2)=\sqrt[3]{2 - 1}=1 \) (option c)
• \( x = -7 \): \( f(-7)=\sqrt[3]{-7 - 1}=\sqrt[3]{-8}=-2 \) (option a)

Answer:

For \( x = 0 \): g. - 1
For \( x = 1 \): f. 0
For \( x = 9 \): b. 2
For \( x = 2 \): c. 1
For \( x = -7 \): a. - 2