Question

select two functions with the same end behaviors of $f(x) = (x - 1)^{\frac{1}{3}} + 2$ as $x$ approaches positive infinity and approaches negative infinity.
$f(x) = -sqrt3{x} + 4$
$f(x) = (x - 1)^3 + 2$
$f(x) = sqrt{x} + 2$
$f(x) = x^2 + 6x + 11$
$f(x) = sqrt3{x + 5} - 7$

Explanation:

Step1: Analyze end behavior of $f(x)$

For $f(x)=(x-1)^{\frac{1}{3}}+2 = \sqrt[3]{x-1}+2$:

• As $x\to+\infty$, $\sqrt[3]{x-1}\to+\infty$, so $f(x)\to+\infty$
• As $x\to-\infty$, $\sqrt[3]{x-1}\to-\infty$, so $f(x)\to-\infty$

This is the end behavior of an odd-powered root function (cube root, odd index).

Step2: Check each option

Option1: $f(x)=-\sqrt[3]{x}+4$

• $x\to+\infty$: $-\sqrt[3]{x}\to-\infty$, $f(x)\to-\infty$ (does not match)

Option2: $f(x)=(x-1)^3+2$

• $x\to+\infty$: $(x-1)^3\to+\infty$, $f(x)\to+\infty$
• $x\to-\infty$: $(x-1)^3\to-\infty$, $f(x)\to-\infty$ (matches)

Option3: $f(x)=\sqrt{x}+2$

• Domain is $x\geq0$, undefined for $x\to-\infty$ (does not match)

Option4: $f(x)=x^2+4x+11$

• $x\to+\infty$ and $x\to-\infty$: $x^2\to+\infty$, $f(x)\to+\infty$ (does not match)

Option5: $f(x)=\sqrt[3]{x+5}-7$

• $x\to+\infty$: $\sqrt[3]{x+5}\to+\infty$, $f(x)\to+\infty$
• $x\to-\infty$: $\sqrt[3]{x+5}\to-\infty$, $f(x)\to-\infty$ (matches)

Answer:

1. $f(x)=(x-1)^3+2$
2. $f(x)=\sqrt[3]{x+5}-7$