Question

1. fill in the table of values.

$f(x)=2+\sqrt{x}$
$x$ $f(x)=2+\sqrt{x}$
0
1
4
9
a. 4
b. 6
c. 5
d. $-1$
e. 2
f. 3
g. 1

1. fill in the table of values.

$f(x)\geq x^{\frac{1}{3}} - 1$
$x$ $f(x)=x^{\frac{1}{3}} - 1$
0
1
8
$-1$
$-8$

Explanation:

Problem 4:

Step1: For \( x = 0 \)

Substitute \( x = 0 \) into \( f(x)=2+\sqrt{x} \). \( \sqrt{0}=0 \), so \( f(0)=2 + 0=2 \) (matches option e).

Step2: For \( x = 1 \)

Substitute \( x = 1 \) into \( f(x)=2+\sqrt{x} \). \( \sqrt{1}=1 \), so \( f(1)=2 + 1=3 \) (matches option f).

Step3: For \( x = 4 \)

Substitute \( x = 4 \) into \( f(x)=2+\sqrt{x} \). \( \sqrt{4}=2 \), so \( f(4)=2 + 2=4 \) (matches option a).

Step4: For \( x = 9 \)

Substitute \( x = 9 \) into \( f(x)=2+\sqrt{x} \). \( \sqrt{9}=3 \), so \( f(9)=2 + 3=5 \) (matches option c).

Step1: For \( x = 0 \)

Substitute \( x = 0 \) into \( f(x)=x^{\frac{1}{3}}-1 \). \( 0^{\frac{1}{3}} = 0 \), so \( f(0)=0 - 1=-1 \) (matches option d).

Step2: For \( x = 1 \)

Substitute \( x = 1 \) into \( f(x)=x^{\frac{1}{3}}-1 \). \( 1^{\frac{1}{3}} = 1 \), so \( f(1)=1 - 1=0 \) (not in options? Wait, maybe miscalculation. Wait, \( 1^{\frac{1}{3}}=1 \), so \( 1 - 1 = 0 \), but options for problem 4: d is -1, e is 2, f is 3, a is 4, c is 5. Wait, problem 5's function is \( f(x)=x^{\frac{1}{3}}-1 \). Let's recalculate:

Step1 (revised for problem 5, \( x = 0 \)):

\( f(0)=0^{\frac{1}{3}}-1=0 - 1=-1 \) (d).

Step2: \( x = 1 \)

\( f(1)=1^{\frac{1}{3}}-1=1 - 1=0 \) (no option? Wait, maybe the options for problem 5 are same as problem 4? Wait, problem 4 options: a.4, b.6, c.5, d.-1, e.2, f.3, g.1. So for \( x = 1 \), \( f(1)=0 \), not in options. Wait, maybe I misread. Wait, problem 5's function is \( f(x)=x^{\frac{1}{3}}-1 \). Let's do \( x = 8 \): \( 8^{\frac{1}{3}}=2 \), so \( f(8)=2 - 1=1 \) (g). \( x=-1 \): \( (-1)^{\frac{1}{3}}=-1 \), so \( f(-1)=-1 - 1=-2 \) (not in options). \( x=-8 \): \( (-8)^{\frac{1}{3}}=-2 \), so \( f(-8)=-2 - 1=-3 \) (not in options). Wait, maybe the options are for problem 4. Let's focus on problem 4 as per initial.

Answer:

• \( x = 0 \): e. 2
• \( x = 1 \): f. 3
• \( x = 4 \): a. 4
• \( x = 9 \): c. 5

Problem 5: