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The function is `f ( x ) = x sqrt ( 4 - x^2 ) . ` It is defined on `[ -2 , 2 ] ` because the expression under the square root must be non-negative. The function is continuous on its domain and is infinitely differentiable inside...

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Hello!

The function is `f ( x ) = x sqrt ( 4 - x^2 ) . ` It is defined on `[ -2 , 2 ] ` because the expression under the square root must be non-negative. The function is continuous on its domain and is infinitely differentiable inside it.

The first derivative is

`f ' ( x ) = sqrt ( 4 - x^2 ) - ( 2 x^2 ) / ( 2 sqrt ( 4 - x^2 ) ) = ( 2 ( 2 - x^2 ) ) / sqrt ( 4 - x^2 ) . `

It does not exist (is infinite) at `x = +- 2 ` and is zero at `x = +- sqrt ( 2 ) . ` It is negative on `(-2, -sqrt(2)), ` positive on `(-sqrt(2),sqrt(2)) ` and negative on `(sqrt(2),2). ` Therefore f(x) has a local minimum at `x=-sqrt(2) ` and a local maximum at `x=sqrt(2).`

The second derivative is

`f''(x) = (2x(x^2-6) ) / ( 4-x^2 )^( 3 / 2 ) ` (check),

which is positive on (-2,0) and negative on (0,2), so f(x) has an inflection point at `x=0.`

There are no asymptotes because neither x nor y tend to infinity.

There are three x-intercepts, `( -2 , 0 ), ` `( 0 , 0 ) ` and `( 2 , 0 ), ` and `( 0 , 0 ) ` is also an y-intercept. Also the function is odd.

The graph is made with Desmos.com.