consider the following function.\n w(x)=\begin{cases}-\frac{2}{5}x & \text{if }x < - 2\\\frac{10}{9}sqrt3{x}&…

Answer:

The function $w(x)=\frac{10}{9}\sqrt[3]{x}$ on the interval $[- 2,\infty)$ is a cube - root function. The general shape of the graph of the cube - root function $y = \sqrt[3]{x}$ is a curve that starts from the third quadrant (when considering the full domain of $y=\sqrt[3]{x}$), passes through the origin, and extends into the first quadrant. For $y=\frac{10}{9}\sqrt[3]{x}$, the coefficient $\frac{10}{9}>0$, so the graph has the same general shape as $y = \sqrt[3]{x}$ and is increasing on the interval $[-2,\infty)$. The graph starts at the point $(-2,\frac{10}{9}\sqrt[3]{-2})=(-2,-\frac{10}{9}\sqrt[3]{2})$ and increases as $x$ increases towards $\infty$.

Explanation:

Step1: Identify the function type

The function on $[-2,\infty)$ is $w(x)=\frac{10}{9}\sqrt[3]{x}$, a cube - root function.

Step2: Analyze the coefficient

The coefficient $\frac{10}{9}>0$, so it is increasing.

Step3: Determine starting point

When $x = - 2$, $w(-2)=\frac{10}{9}\sqrt[3]{-2}=-\frac{10}{9}\sqrt[3]{2}$, and it increases as $x$ goes to $\infty$.