Question

1. choose the correct equation or inequality by using the indicated points.

$f(x)=1+\sqrt{x}$
$f(x)=x^3 + 1$
$f(x)=1 - \sqrt3{x}$
$f(x)=1 + \sqrt3{x}$

Explanation:

Step1: Test point (-1,0) in options

Option 1: $f(x)=1+\sqrt{x}$

$\sqrt{-1}$ is not real, so invalid.

Option 2: $f(x)=x^3+1$

$(-1)^3 + 1 = -1 + 1 = 0$, matches.

Option 3: $f(x)=1-\sqrt[3]{x}$

$1 - \sqrt[3]{-1} = 1 - (-1) = 2
eq 0$, invalid.

Option 4: $f(x)=1+\sqrt[3]{x}$

$1 + \sqrt[3]{-1} = 1 + (-1) = 0$, matches.

Step2: Test point (1,2) in remaining options

Option 2: $f(x)=x^3+1$

$1^3 + 1 = 1 + 1 = 2$, matches.

Option 4: $f(x)=1+\sqrt[3]{x}$

$1 + \sqrt[3]{1} = 1 + 1 = 2$, matches.

Step3: Test point (0,1) in remaining options

Option 2: $f(x)=x^3+1$

$0^3 + 1 = 0 + 1 = 1$, matches.

Option 4: $f(x)=1+\sqrt[3]{x}$

$1 + \sqrt[3]{0} = 1 + 0 = 1$, matches.

Step4: Check curve shape

The graph is a cube root curve (smooth, defined for all $x$, increasing slowly), not a cubic curve (which has a steeper, S-shaped slope). $f(x)=1+\sqrt[3]{x}$ is a shifted cube root function, matching the graph's shape.

Answer:

$\boldsymbol{f(x)=1+\sqrt[3]{x}}$ (the fourth option)