Question

1. fill in the table of values.

$f(x) \geq x^{\frac{1}{3}} - 1$
$x$ $f(x) = x^{\frac{1}{3}} - 1$
0 e
1 f
8
-1
-8
a. 1
b. -4
c. -2
d. 3
e. 2
f. -3
g. 0
h. -1

Explanation:

Step1: For \( x = 0 \)

Substitute \( x = 0 \) into \( f(x)=x^{\frac{1}{3}} - 1 \). We know that \( 0^{\frac{1}{3}}=0 \), so \( f(0)=0 - 1=-1 \). Wait, but the option for e? Wait, maybe I misread. Wait, the function is \( f(x)=x^{\frac{1}{3}}-1 \). Let's recalculate:

Wait, \( x^{\frac{1}{3}} \) is the cube root. So for \( x = 0 \): \( \sqrt[3]{0}-1=0 - 1=-1 \), which is option h.

Step2: For \( x = 1 \)

Substitute \( x = 1 \) into \( f(x)=x^{\frac{1}{3}}-1 \). \( \sqrt[3]{1}-1 = 1 - 1=0 \), which is option g.

Step3: For \( x = 8 \)

Substitute \( x = 8 \) into \( f(x)=x^{\frac{1}{3}}-1 \). \( \sqrt[3]{8}-1=2 - 1 = 1 \), which is option a.

Step4: For \( x=-1 \)

Substitute \( x = -1 \) into \( f(x)=x^{\frac{1}{3}}-1 \). \( \sqrt[3]{-1}-1=-1 - 1=-2 \), which is option c.

Step5: For \( x = -8 \)

Substitute \( x=-8 \) into \( f(x)=x^{\frac{1}{3}}-1 \). \( \sqrt[3]{-8}-1=-2 - 1=-3 \), which is option f. Wait, but the original table has e for x=0, f for x=1. Wait, maybe the labels are for each dropdown. Let's re - check the table:

For \( x = 0 \), the dropdown is labeled e. Let's recalculate \( x = 0 \): \( 0^{\frac{1}{3}}-1=-1 \) (option h). But maybe the problem's options are mixed? Wait, no, let's do each step carefully.

Wait, the function is \( f(x)=x^{\frac{1}{3}}-1 \). Let's compute each:

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

But the table has for \( x = 0 \) the dropdown labeled e, \( x = 1 \) labeled f. Wait, maybe the labels are e, f, etc., and we need to match the values. Let's list the values:

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

But the problem's table has e for x=0, f for x=1. Maybe the initial labels are e, f, etc., and we need to assign the correct options. Let's re - express:

For \( x = 0 \): value is -1 (h), so e should be h.

For \( x = 1 \): value is 0 (g), so f should be g.

For \( x = 8 \): value is 1 (a), so the third dropdown (x=8) is a.

For \( x=-1 \): value is -2 (c), so the fourth dropdown is c.

For \( x=-8 \): value is -3 (f), so the fifth dropdown is f.

But the question is to fill in the table. Let's present each:

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

But the original table has e for x=0, f for x=1. So:

• e (x=0): h. -1
• f (x=1): g. 0
• x=8: a. 1
• x=-1: c. -2
• x=-8: f. -3

Answer:

• For \( x = 0 \) (e): h. -1
• For \( x = 1 \) (f): g. 0
• For \( x = 8 \): a. 1
• For \( x=-1 \): c. -2
• For \( x=-8 \): f. -3