<s>[INST] Range of function $f(x)=x \sqrt{1-x^2}$ How to find range of this function, it looks easy but somehow I am not able to do that .  What is more important to me is , why is wolframalpha unable to find the range , is it not a simple enough  function ?

 [/INST] $1-x^2=(1-x)(1+x)$ needs to be positive because it is under the root. Hence,
$$(1-x)(1+x)\geq 0.$$
What can you conclude for possible values of $x$? The possible values of $x$ are limited on the interval $[-1,1]$. We have $f(-1)=0$ and $f(+1)=0$.
Then consider the derivative of the function
$$f'(x) = \sqrt{1-x^2}+x\dfrac{1}{2\sqrt{1-x^2}}\left(-2x \right)$$
$$=\dfrac{2(1-x^2)-2x^2}{2\sqrt{1-x^2}}=\dfrac{2(1-2x^2)}{2\sqrt{1-x^2}}$$
$$=\dfrac{1-2x^2}{\sqrt{1-x^2}}.$$
The derivative is $0$ for $1-2x^2=0\implies x = \pm \sqrt{1/2}$. The derivative is negative for $1\geq|x|\geq\sqrt{1/2}$ and positive for $|x|\leq \sqrt{1/2}$.
The extremal values are $f(\pm \sqrt{1/2})=\pm\sqrt{1/2}\sqrt{1-1/2}=\pm\sqrt{1/2}\sqrt{1/2}=\pm 0.5.$
As the continuous function vanishes on the bounds of $[-1,1]$ we know that the extremal values of the function will dictate the range of the function. The range is $[-1/2,1/2]$.
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