Question

example 3: graph the function $-2\sqrt3{x + 5}$ and identify the domain and range.

Explanation:

Step1: Analyze the function type

The function is \( y = -2\sqrt[3]{x + 5} \). The cube root function \( \sqrt[3]{u} \) is defined for all real numbers \( u \), so for \( u=x + 5 \), \( x+5 \) can be any real number, which means \( x \) can be any real number.

Step2: Determine the domain

Since there are no restrictions on \( x \) (the cube root is defined for all real numbers), the domain of the function \( y=-2\sqrt[3]{x + 5} \) is all real numbers, i.e., \( (-\infty,\infty) \).

Step3: Analyze the range

The cube root function \( \sqrt[3]{x + 5} \) can take any real value (because as \( x\to\infty \), \( \sqrt[3]{x + 5}\to\infty \); as \( x\to-\infty \), \( \sqrt[3]{x + 5}\to-\infty \)). When we multiply by - 2, the function \( - 2\sqrt[3]{x + 5} \) can also take any real value (multiplying a real number by - 2 still gives a real number, and the range of \( \sqrt[3]{x+5} \) being \( (-\infty,\infty) \) implies the range of \( - 2\sqrt[3]{x + 5} \) is also \( (-\infty,\infty) \)).

Step4: Graphing the function (brief idea)

The parent function is \( y=\sqrt[3]{x} \). First, we have a horizontal shift: \( y = \sqrt[3]{x+5} \) is the parent function shifted 5 units to the left. Then, we have a vertical stretch by a factor of 2 and a reflection over the x - axis (because of the - 2 coefficient) to get \( y=-2\sqrt[3]{x + 5} \). We can plot key points: when \( x=-5 \), \( y = 0 \); when \( x=-5 + 1=-4 \), \( y=-2\sqrt[3]{1}=-2 \); when \( x=-5 - 1=-6 \), \( y=-2\sqrt[3]{-1}=2 \); when \( x=-5 + 8 = 3 \), \( y=-2\sqrt[3]{8}=-4 \); when \( x=-5-8=-13 \), \( y=-2\sqrt[3]{-8}=4 \). Then we connect these points with a smooth curve.

Answer:

• Domain: \( (-\infty,\infty) \)
• Range: \( (-\infty,\infty) \)
• Graph: A smooth curve (cubic - like) shifted 5 units left, stretched vertically by a factor of 2, and reflected over the x - axis. Key points can be used to sketch it as described above.