Problem 111 - Precalculus Essentials

Question

Begin by graphing the cube root function, \(f(x)=\sqrt[3]{x} .\) Then use transformations of this graph to graph the given function. $$h(x)=\frac{1}{2} \sqrt[3]{x+2}$$

Step-by-Step Solution

Answer

The graph of the function \(h(x)=\frac{1}{2} \sqrt[3]{x+2}\) is a transformation of the original cube root function \(f(x)=\sqrt[3]{x}\), where it is shifted two units to the left and vertically compressed by a factor of \(\frac{1}{2}\).

Step 1: Graph the basic function

Begin by graphing the basic cube root function: \(f(x)=\sqrt[3]{x}\). Remember, the cube root function starts from negative infinity, crosses the origin, and then continues to positive infinity.

Step 2: Recognize the transformations

Identify the transformations present in the given function, \(h(x)=\frac{1}{2} \sqrt[3]{x+2}\). There are two transformations here: a horizontal shift (or translation) to the left by 2 units due to the '+2' inside the cube root, and a vertical dilation by a factor of \(\frac{1}{2}\) due to the multiplier outside the cube root.

Step 3: Apply the horizontal shift

First, apply the horizontal shift by moving every point of the original function \(f(x)=\sqrt[3]{x}\) two units to the left. This is because of the '+2' inside the cube root.

Step 4: Apply the vertical dilation

Then, apply the vertical dilation by reducing the y-value of every point by half. This is due to the multiplier of \(\frac{1}{2}\) outside the cube root.

Step 5: Finalize the graph

Now, the transformed function \(h(x)=\frac{1}{2} \sqrt[3]{x+2}\) should be graphed. It should look like the original function \(f(x)=\sqrt[3]{x}\) that has been shifted two units to the left and vertically compressed by a factor of \(\frac{1}{2}\). The graph will start from negative infinity, reach a minimum at (-2,0), and then continue to positive infinity.