Question

consider the following function.
p(x)=left{\begin{array}{ll}\frac{2}{x^{3}} & \text { if } x<3 \\ -\frac{8}{7} x & \text { if } x>3end{array}
ight.
step 1 of 3: identify the general shape and direction of the graph of this function on the interval ((-infty, 3)).

Explanation:

Step1: Analyze the function for \(x < 3\)

The function for \(x<3\) is \(y = \frac{2}{x^{3}}=2x^{- 3}\).

Step2: Recall the properties of power functions

For a power function \(y = ax^{n}\), when \(n=-3<0\) and \(a = 2>0\).
As \(x\to-\infty\), \(y = 2x^{-3}=\frac{2}{x^{3}}\to0\) (approaching \(0\) from the negative side since \(x^{3}<0\) for \(x<0\)). As \(x\) approaches \(0\) from the negative side (\(x\to0^{-}\)), \(y=\frac{2}{x^{3}}\to-\infty\). As \(x\) approaches \(0\) from the positive side (\(x\to0^{+}\)) and \(x<3\), \(y = \frac{2}{x^{3}}\to+\infty\). As \(x\) approaches \(3\) from the left side (\(x\to3^{-}\)), \(y=\frac{2}{3^{3}}=\frac{2}{27}\).

The general shape of \(y = \frac{2}{x^{3}}\) is a hyper - bolic - like curve. Since the exponent \(n=-3\) (odd), the function is symmetric about the origin. For \(x<3\), when \(x>0\) and \(x < 3\), as \(x\) increases, \(y=\frac{2}{x^{3}}\) decreases. When \(x<0\), as \(x\) increases (towards \(0\)), \(y=\frac{2}{x^{3}}\) increases (from \(-\infty\) towards \(0\)).

Answer:

The graph of \(y = p(x)\) for \(x\in(-\infty,3)\) has a hyper - bolic - like shape. For \(x < 0\), as \(x\) increases (towards \(0\)), the function values increase (from \(-\infty\) towards \(0\)). For \(0