Calculus Question. I am having trouble on this curve sketching problem.

[jamesc96x]

Curve sketching question : Given the equation y= x*(sqrt(4-x^2))


Find x and y intercepts,
Maximum and minimum points
Critical points, and points of inflection
Find the intervals to graph the equation. . For example, I know the domain as -2 < x < 2 Help would so much appreciated!

 

[HallsofIvy]

Do you not know the meanings of these words?

The "y-intercept" is where the graph crosses (intercepts) the y-axis. And at the y-axis, x= 0. The x- coordinate of the y-intercept is, of course, 0, and the y- coordinate is \(\displaystyle x\sqrt{4- x^2})[/tex[ with x= 0. That should be easy to find.

The "x-intercept" is where the graph crosses the x-axis. And at the x-axis, y= 0. So you need to solve the equation \(\displaystyle x\sqrt{4- x^2}= 0\). That will have three distinct values.

The "critical points" are points where the derivative either is 0 or does not exist. To differentiate \(\displaystyle x(4- x^2)^{1/2}\) use the product rule and the chain rule.

A maximum point will be where the first derivative is 0 and the second derivative is negative. A minimum where the first derivative is 0 and the second derivative is positive.

Points of inflection are points where the second derivative changes sign.\)

 

[jamesc96x]

I found the first derivative, its: y'= (4-2x^2)/√(4-x^2)
I also got the second derivative. Its:
y"=(2x^3-12x)/(4-x^2)^(3/2)

I know that the critical points are √2 and
-√2. I also found that the inflection points are x= 0 x , = -√6, and x= √6.

Since the domain is -2 <= x <= 2, do I disregard the -√6 and √6? √6 = 2.4.

I'm having a lot of trouble putting this together and creating a chart and graphing all of this.