Range of function $f(x)=x \sqrt{1-x^2}$

Question

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 ?

Answer 1

As $\sqrt{1-x^2}\ge0,1-x^2\ge0\implies-1\le x\le1$

Method $\#1:$

Using Arithmetic Mean-Geometric Mean Inequality,

$$\dfrac{x^2+1-x^2}2\ge \sqrt{x^2(1-x^2)}=|x|\sqrt{1-x^2}$$

If $x\le0, -x\sqrt{1-x^2}\le\dfrac12\iff x\sqrt{1-x^2}\ge-\dfrac12$

What if $x>0?$

Method $\#2:$

WLOG $x=\sin y,-\dfrac\pi2\le y\le\dfrac\pi2\implies\cos y=\sqrt{1-x^2}\ge0$

as $x\sqrt{1-x^2}=\dfrac{\sin2y}2$

Answer 2

$1-x^2=(1-x)(1+x)$ 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 $[-1/2,1/2]$.

Answer 3

It is $$-\frac{1}{2}\le y\le \frac{1}{2}$$ this can be obtained: $$x\sqrt{1-x^2}\le \frac{1}{2}$$ $$4x^2(1-x^2)\le 1$$ or $$0\le(2x-1)^2$$

Answer 4

Since the domain is $[-1,1]$ (otherwise the square root is not defined), we only have to take $x$ in that domain. Now, the first factor take values in $[-1,1]$, instead the second one in $[0,1]$, but it is not in general sufficient to deduce the range of the product. You have to compute the first derivative and find the maxima and minima of the function, and evaluate it on them. Then, by continuity of the function, it will take all the values between the absolute minimum and the absolute maximum, and that will be the range (just to be technical, it is what Darboux' theorem says).

Answer 5

Domain of $f$ is $[-1,1]$ since $1-x²≥0$ because it is under square root. The function is continuous so find critical points using $\frac{df}{dx} = 0$ and then find the value of function at these critical points and also at the boundary points i.e., $x={-1,1}$. Then your range will be from minimum value among these values to maximum value among these values.