Question

the graph of $f(x) = \sqrt3{x + 8}$ is shown.
which statement is true?
\bigcirc the function is only increasing when $x \geq -8$
\bigcirc the function is only increasing when $x \geq 0$
\bigcirc the function is always decreasing
\bigcirc the function is always increasing

Explanation:

Step1: Analyze the function type

The function is \( f(x)=\sqrt[3]{x + 8}\), which is a cube - root function. The general form of a cube - root function is \(y = \sqrt[3]{x - h}+k\), and its graph is a smooth curve. For the cube - root function \(y=\sqrt[3]{u}\), where \(u=x + 8\), we know that the derivative of \(y = \sqrt[3]{u}\) with respect to \(u\) is \(y^\prime=\frac{1}{3}u^{-\frac{2}{3}}=\frac{1}{3(u^{\frac{2}{3}})}\), and the derivative of \(u=x + 8\) with respect to \(x\) is \(u^\prime = 1\). By the chain rule, the derivative of \(y=\sqrt[3]{x + 8}\) with respect to \(x\) is \(f^\prime(x)=\frac{1}{3((x + 8)^{\frac{2}{3}})}\).

Since \((x + 8)^{\frac{2}{3}}=\sqrt[3]{(x + 8)^{2}}\geq0\), and \((x + 8)^{2}=0\) when \(x=-8\), so \(f^\prime(x)=\frac{1}{3((x + 8)^{\frac{2}{3}})}\geq0\) for all real \(x\) (and \(f^\prime(x)>0\) when \(x
eq - 8\), \(f^\prime(-8)\) is undefined but the function is continuous at \(x = - 8\)).

Step2: Analyze the graph

Looking at the given graph of \(f(x)=\sqrt[3]{x + 8}\), as we move from left to right (as \(x\) increases), the \(y\) - value of the function is always increasing. For example, when \(x\) is a smaller number (more negative), the \(y\) - value is lower, and as \(x\) increases (becomes less negative, then positive), the \(y\) - value keeps increasing.

Now let's analyze each option:

• Option 1: "The function is only increasing when \(x\geq - 8\)": The function is increasing for all real \(x\), not just \(x\geq - 8\) (it is increasing for \(x < - 8\) as well).
• Option 2: "The function is only increasing when \(x\geq0\)": The function is increasing for \(x<0\) (e.g., when \(x=-9\), \(f(-9)=\sqrt[3]{-9 + 8}=\sqrt[3]{-1}=-1\); when \(x=-8\), \(f(-8)=\sqrt[3]{-8 + 8}=0\); when \(x=-7\), \(f(-7)=\sqrt[3]{-7 + 8}=1\), so as \(x\) increases from - 9 to - 7, \(f(x)\) increases). So this option is wrong.
• Option 3: "The function is always decreasing": The graph shows that as \(x\) increases, \(y\) increases, so the function is not decreasing.
• Option 4: "The function is always increasing": From the derivative analysis and the graph analysis, as \(x\) increases, the function value \(f(x)\) always increases.

Answer:

The function is always increasing (the fourth option: The function is always increasing)