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 ? • Type range of x * sqrt(1-x^2) Mar 17 '19 at 9:53
• I put the same text , still doesn't come . Mar 17 '19 at 10:06
• It comes for me Mar 17 '19 at 10:06

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

Method $$\#1:$$

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

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

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.

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

• x lies in [-1,1] Mar 17 '19 at 10:01
• Very good, what is $f(-1)$ and $f(+1)$? And calculate the derivative and set it to $0$. What do you get? What is the extremal point of your function? Mar 17 '19 at 10:03
• + and - 1/√2 as x coordinates and therefore range is [-1/2,1/2] , so using differentiation is the standard approach for such functions, are there other methods ? thanks Mar 17 '19 at 10:24
• The procedure for standard problems with a continuous function is given by the following method. First, look if your domain is bounded. If yes evaluate the function value at the bounds. If no evaluate the function or $|x|\to \infty$. Then calculate the derivative and set it equal to $0$ to obtain extremal values of your function. Then look at which of the determined points is the maximal point (upper bound for range) and which is minimal (lower bound of range). Mar 17 '19 at 10:38

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 $$0\le(2x-1)^2$$

• How did you get x√ (1-x^2) less than equal 1/2 ? Mar 17 '19 at 10:09
• This is the range, written above by me Mar 17 '19 at 10:11
• And the next lines are the proof of this fact Mar 17 '19 at 10:11

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).