which cube root function is always decreasing as x increases?\n$f(x)=sqrt3{x - 8}$\n$f(x)=sqrt3{x}-5$\n$f(x)=…

Explanation:

Step1: Recall cube - root function property

The basic cube - root function $y = \sqrt[3]{x}$ is an increasing function. That is, as $x$ increases, $y$ increases. For a function of the form $y=a\sqrt[3]{x - h}+k$, the sign of $a$ determines the direction of the function's increase or decrease.

Step2: Analyze each option

• Option 1: $f(x)=\sqrt[3]{x - 8}$. Here $a = 1$, and since $a>0$, as $x$ increases, $x-8$ increases and $\sqrt[3]{x - 8}$ increases.
• Option 2: $f(x)=\sqrt[3]{x}-5$. Here $a = 1$, and as $x$ increases, $\sqrt[3]{x}$ increases, so $\sqrt[3]{x}-5$ increases.
• Option 3: $f(x)=\sqrt[3]{-(5 - x)}=\sqrt[3]{x - 5}$. Here $a = 1$, and as $x$ increases, $x - 5$ increases and $\sqrt[3]{x - 5}$ increases.
• Option 4: $f(x)=-\sqrt[3]{x}+5$. Here $a=-1$. When $x$ increases, $\sqrt[3]{x}$ increases, but since there is a negative sign in front of $\sqrt[3]{x}$, $-\sqrt[3]{x}$ decreases, and $-\sqrt[3]{x}+5$ also decreases.

Answer:

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