Finding range if we know domain in interval form

Question

I was required to solve for the domain and range of the function, $$y=\sqrt{x-x^2}$$

I find the domain to be $[0,1]$.

Now that I have the domain in this form, how can I find the range in the interval form?

Answer 1

$x-x^2 = -(x^2-x +\frac14 - \frac14) = \frac14-\left(x-\frac12\right)^2$

So, $\sqrt{x-x^2} = \sqrt{\frac14-\left(x-\frac12\right)^2}$

$0\le x\le1 \iff -\frac12\le x-\frac12 \le \frac12\iff0\le\left(x-\frac12\right)^2\le\frac14 \iff 0\le\frac14 - \left(x-\frac12\right)^2 \le\frac14$

Thus, $0\le\sqrt{\frac14-\left(x-\frac12\right)^2}\le\frac12$. The range is $[0,\frac12]$

Answer 2

Note that $y$ can not be negative. The square root function is always positive. Let us call this CONDITION I.

Also, the given equation can be rearranged as follows:

$y=\sqrt{x-x^2}\implies y^2=x-x^2 \implies x^2-x+y^2=0$

We complete the square by adding $\frac{1}{4}$ to both sides, we get:

$x^2-x+\frac{1}{4}+y^2=\frac{1}{2}^2 \implies (x-\frac{1}{2})^2+y^2=\frac{1}{2}^2$

Note that this is the equation of a cirlce whose centre lies on the $x$-axis and whose radius is $\frac{1}{2}$. Therefore $y\le \frac{1}{2}$. Let us call this CONDITION II.

Putting CONDITION I and CONDITION II together, we get:

$$0 \le y \le \frac{1}{2}$$

Answer 3

The radicand $f(x)=x-x^2$ satisfies $f(0)=f(1)=0$. And, for $0\leq x \leq 1$ we have $0 \leq x^2 \leq x \leq 1$ so $f(x)\geq 0$ on $[0,1]$. We just need to find its maximum on $[0,1]$ to then work out the maximum $y$. Well, $f'(x)=1-2x=0$ means $x=\frac{1}{2}$ and $f''(x)=-2<0$ means it is a maximum. And, at this point, we have maximum $y=\sqrt{f(\frac{1}{2})}=\frac{1}{2}$. Since the function is continuous it takes all values between $0$ and $\frac{1}{2}$ so its range is $[0,\frac{1}{2}]$.