Question

a. find the open interval(s) on which the function is increasing and decreasing
b. identify the functions local and absolute extreme values, if any, saying where they occur.
g(x)=x\sqrt{18 - x^{2}}
a. on what open interval(s), if any, is the function increasing? select the correct choice below and,
if necessary, fill in the answer box(es) to complete your choice.
○ a. the function g is increasing on the open interval(s)
(type your answer in interval notation. type an exact answer in simplified form. use a comma to
separate answers as needed.)
○ b. the function is never increasing

Explanation:

Step1: Find the domain and derivative

• Domain:

For the function \(y = g(x)=x\sqrt{18 - x^{2}}\), the expression under the square - root must be non - negative. So, \(18-x^{2}\geq0\), which gives \((x + 3\sqrt{2})(x - 3\sqrt{2})\leq0\). The domain is \([-3\sqrt{2},3\sqrt{2}]\).

• Derivative:

Using the product rule \((uv)^\prime=u^\prime v+uv^\prime\), where \(u = x\) and \(v=\sqrt{18 - x^{2}}=(18 - x^{2})^{\frac{1}{2}}\).
\(u^\prime = 1\) and \(v^\prime=\frac{1}{2}(18 - x^{2})^{-\frac{1}{2}}\times(-2x)=\frac{-x}{\sqrt{18 - x^{2}}}\).
Then \(g^\prime(x)=\sqrt{18 - x^{2}}+x\times\frac{-x}{\sqrt{18 - x^{2}}}=\frac{18 - x^{2}-x^{2}}{\sqrt{18 - x^{2}}}=\frac{18 - 2x^{2}}{\sqrt{18 - x^{2}}}\).
Set \(g^\prime(x)=0\), then \(18 - 2x^{2}=0\).
\(2x^{2}=18\), \(x^{2}=9\), \(x=\pm3\).

Step2: Determine the sign of the derivative

• Test intervals:

We consider the intervals \((-3\sqrt{2},-3)\), \((-3,3)\) and \((3,3\sqrt{2})\).
For \(x\in(-3\sqrt{2},-3)\), let \(x=-4\) (since \(-3\sqrt{2}\approx - 4.24\)). Then \(g^\prime(-4)=\frac{18-2\times(-4)^{2}}{\sqrt{18-(-4)^{2}}}=\frac{18 - 32}{\sqrt{18 - 16}}=\frac{-14}{\sqrt{2}}<0\).
For \(x\in(-3,3)\), let \(x = 0\). Then \(g^\prime(0)=\frac{18-2\times0^{2}}{\sqrt{18-0^{2}}}=\frac{18}{\sqrt{18}}=\sqrt{18}>0\).
For \(x\in(3,3\sqrt{2})\), let \(x = 4\) (since \(3\sqrt{2}\approx4.24\)). Then \(g^\prime(4)=\frac{18-2\times4^{2}}{\sqrt{18 - 4^{2}}}=\frac{18 - 32}{\sqrt{18 - 16}}=\frac{-14}{\sqrt{2}}<0\).

Answer:

A. The function \(g\) is increasing on the open interval \((-3,3)\)