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

Question

I have to find the range of $f(x)=x\sqrt{1-x^2}$ on the interval $[-1,1]$. I have done so by setting $x=\sin\theta$ and thus finding it to be $[-0.5,0.5]$.

Let $x=\sinθ$. Then, for $x\in[-1,1]$ we get that $θ \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Thus, $f(x)$ becomes:

$f(\theta)=\sin\theta \sqrt{1-(\sin\theta)^2}= \sin\theta |\cos\theta|$. Since for $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ we have that $cos\theta \geq 0$:

$f(θ)=\sinθ\cosθ=\frac{1}{2}\sin2θ$. Which has maximum value $\frac{1}{2}$ when $\theta = \frac{\pi}{4}$ and minimum value $-\frac{1}{2}$ when $\theta = -\frac{π}{4}$. When $\theta = \frac{\pi}{4}$ we have $x=\frac{\sqrt{2}}{2}$ and when $\theta = -\frac{\pi}{4}$ we have $x=-\frac{\sqrt{2}}{2}$, which are the maximum and minimum positions respectively. So, $f(\frac{\sqrt{2}}{2})=\frac{1}{2}$ and $f(-\frac{\sqrt{2}}{2})=-\frac{1}{2}$. Thus the range of $f(x)$ is $[-\frac{1}{2},\frac{1}{2}]$.

I know that it can be found with derivatives as well. I was wondering how can quadratic theory can be used to find the Range.

Answer 1

You could reason like this: $f(x)$ will have its maximum value at the same point as $g(x) = (f(x))^2 = x^2 - x^4.$ This is quadratic in $x^2$ and an even function. Since $z-z^2$ takes its maximum value when $z=1/2$ (by finding the vertex of the parabola) then we know $g(x)$, and hence $f(x)$ takes its maximum value when $x^2=1/2$ or $x=\pm 1/\sqrt{2}.$ And $f(1/\sqrt{2}) = 1/2.$ To get the lower end of the range, you can argue by symmetry.

Answer 2

Alternatively, observe that the function is odd over $[-1,1]$. Hence, if the max value occurs at $x_0 > 0$, then the min occurs at $-x_0$. But if $x > 0$, then by AM-GM inequality: $f(x) =x\sqrt{1-x^2} \le \dfrac{x^2+ (1-x^2)}{2}=\dfrac{1}{2}$ with $=$ occurs when $x=\sqrt{1-x^2}\implies x^2 = 1-x^2\implies x=\dfrac{1}{\sqrt{2}}$. So the max is $\dfrac{1}{2}$ and the min is $-\dfrac{1}{2}$.