<s>[INST] Inspecting the function $f(x)=-x\sqrt{1-x^2}$ We are just wrapping up the first semester calculus with drawing graphs of functions. I sometimes feel like my reasoning is a bit shady when I am doing that, so I decided to ask you people from Math.SE. 
I am supposed to draw a graph (and show my working) of the function $f(x)=-x\sqrt{1-x^2}$: Below is my work, I'd be very grateful for any comments on possible loopholes in my reasoning (the results should be correct), thanks!

$$\text{1. } f(x)=-x\sqrt{1-x^2}$$
Domain: We have a square-root function, therefore we need $1-x^2\geq0$. From that we get$x\in[-1,1]$. As this is the only necessary condition, $D(f)=[-1,1]$. The function is also continuous on this interval as there is nothing that would produce a discontinuity.
Symmetry: $f(x)$ is an odd function, as $-f(x)=f(-x)$, as demonstrated here:
$$-f(x)=f(-x)$$
$$-(-x\sqrt{1-x^2})=-(-x)\sqrt{1-(-x)^2}$$
$$x\sqrt{1-x^2}=x\sqrt{1-x^2}$$
Therefore we are only interested in the interval $[0,1]$, as the function will behave symmetrically on $[-1,0]$.
x and y intercepts: $f(x)=0$ holds when $x=0,1$, those are then the x-intercepts. Thus also the y-intercept is at $(0,0)$
First derivative
$$\begin{align}
\ f'(x) &=(-x\sqrt{1-x^2})' \\
 & = (-x)'\sqrt{1-x^2}+(-x)(\sqrt{1-x^2})' \\ 
 & = -\sqrt{1-x^2}+(-x)\frac{1}{2\sqrt{1-x^2}}(-2x) \\ 
 & = -\frac{1-x^2}{\sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}} \\ 
 & = \frac{2x^2-1}{\sqrt{1-x^2}} \\ 
\end{align}$$
Local minima and maxima: The equality $\frac{2x^2-1}{\sqrt{1-x^2}}=0$ yields $x=\frac{1}{\sqrt2}$. As $f(\frac{1}{\sqrt2})=-\frac{1}{2}$ and because we know the x-intercepts, we can conclude that this is the local minimum and the function is decreasing at $x\in [0,\frac{1}{\sqrt2})$ and increasing at $x\in(\frac{1}{\sqrt2},1]$. That can also be concluded from the fact that $f′≤0$ on the interval $[0,\frac{1}{\sqrt2}]$ and $f′≥0$ on the interval $[\frac{1}{\sqrt2},1]$.
Second derivative 
$$\begin{align}
\ f''(x) &=(\frac{2x^2-1}{\sqrt{1-x^2}})' \\
 & = \frac{(2x^2-1)'\sqrt{1-x^2}-(2x^2-1)(\sqrt{1-x^2})'}{1-x^2} \\ 
 & = \frac{4x\sqrt{1-x^2}-(2x^2-1)\frac{-x}{\sqrt{1-x^2}}}{1-x^2} \\ 
 & = \frac{4x\frac{1-x^2}{\sqrt{1-x^2}}-(2x^2-1)\frac{-x}{\sqrt{1-x^2}}}{1-x^2} \\ 
 & = \frac{\frac{-2x^3+3x}{\sqrt{1-x^2}}}{1-x^2} \\ 
 & = \frac{-2x^3+3x}{\sqrt{(1-x^2)^3}} \\ 
\end{align}$$
Inflection point and concavity: $f''(x)=0$ has only one result and that is $x=0$. If we look at the inequality for $x\in(0,1]$.
$$\begin{align}
\ 0&<f''(x) \\
0&<\frac{-2x^3+3x}{\sqrt{(1-x^2)^3}} \\
0&<-2x^3+3x \\
2x^3&<3x \\
2x^2&<3 \\
x^2&<3/2 \\
\end{align}$$
We can see that $x^2<3/2$ is satisfied on $x\in(0,1]$, thus the function is convex on this interval.
Conclusion: Using the symmetricity of the function, we can conclude that the function is increasing on $[-1,-\frac{1}{\sqrt2}]$ and $[\frac{1}{\sqrt2},1]$ and decreasing on $[-\frac{1}{\sqrt2},\frac{1}{\sqrt2}]$. It has local minimum at $x=\frac{1}{\sqrt2}$ and local maximum at $x=-\frac{1}{\sqrt2}$. It is concave on $[-1,0]$ and convex on $[0,1]$, with point of inflection at $x=0$.
 [/INST] Your answer and/or analysis of the function  $$f(x)=-x\sqrt{1-x^2}$$ is accurate, thorough, well-justified, and consistent with its graph: $\quad \quad f(x)=-x\sqrt{1-x^2}$.  
  $\quad$Source: Wolfram Alpha.

Kudos for the effort you've shown and your accurate and detailed analysis! (Just don't forget to graph the function too!)

</s>