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{50 - 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 separate answers as needed.)
b. the function is never increasing

Explanation:

Step1: Determine the domain

For the function \(g(x)=x\sqrt{50 - x^{2}}\), the expression under the square - root must be non - negative. So, \(50-x^{2}\geq0\), which gives \(x^{2}\leq50\) or \(x\in[-\sqrt{50},\sqrt{50}]=[-5\sqrt{2},5\sqrt{2}]\).

Step2: Find the derivative

Using the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x\) and \(v=\sqrt{50 - x^{2}}=(50 - x^{2})^{\frac{1}{2}}\).
\(u^\prime=1\) and \(v^\prime=\frac{1}{2}(50 - x^{2})^{-\frac{1}{2}}\times(-2x)=\frac{-x}{\sqrt{50 - x^{2}}}\).
Then \(g^\prime(x)=\sqrt{50 - x^{2}}+x\times\frac{-x}{\sqrt{50 - x^{2}}}=\frac{50 - x^{2}-x^{2}}{\sqrt{50 - x^{2}}}=\frac{50 - 2x^{2}}{\sqrt{50 - x^{2}}}\).

Step3: Find critical points

Set \(g^\prime(x) = 0\), so \(\frac{50 - 2x^{2}}{\sqrt{50 - x^{2}}}=0\). Since the denominator \(\sqrt{50 - x^{2}}>0\) for \(x\in(-5\sqrt{2},5\sqrt{2})\), we solve \(50 - 2x^{2}=0\).
\(2x^{2}=50\), \(x^{2} = 25\), \(x=\pm5\).

Step4: Test intervals

• For the interval \((-5\sqrt{2},- 5)\), let \(x=-6\) (but \(x=-6

otin(-5\sqrt{2},-5)\) since \(-5\sqrt{2}\approx - 7.07\)). Let's use a test point \(x=-6\) (invalid, instead take \(x=-6\) is wrong. Let's take \(x=-6\) is wrong. Let's take \(x = - 6\) (wrong approach). Let's use the sign of \(g^\prime(x)\) formula.
The derivative \(g^\prime(x)=\frac{50 - 2x^{2}}{\sqrt{50 - x^{2}}}\).
For \(x\in(-5\sqrt{2},-5)\), pick \(x=-6\) (invalid, correct way: consider the function \(y = 50-2x^{2}\), when \(x\in(-5\sqrt{2},-5)\), \(y = 50-2x^{2}<0\) (because \(x^{2}>25\) for \(|x|>5\)).
For \(x\in(-5,5)\), pick \(x = 0\), then \(g^\prime(0)=\frac{50-0}{\sqrt{50 - 0}}=\sqrt{50}>0\).
For \(x\in(5,5\sqrt{2})\), pick \(x = 6\) (invalid, consider \(y = 50-2x^{2}\), when \(x\in(5,5\sqrt{2})\), \(x^{2}>25\), so \(y = 50 - 2x^{2}<0\).

Answer:

A. The function \(g\) is increasing on the open interval\((-5,5)\)