A question about domains and ranges: find the domain and range of the function $f(x)=\sqrt{1-x^2}$, without any digital help

Question

If $f(x)=\sqrt{1-x^2}$, then the domain of $f$ is $[-1,1]$. Respectively, the range of $f$ is $[0,1]$.

How do I understand that the domain of $f$ is $[-1,1]$ and that the range of $f$ is $[0,1]$, without a graphing calculator?

Let us suppose that I don't know the given intervals from start. Instead, suppose that I know only that the function that I should find the domain and range of is $f(x)=\sqrt{1-x^2}$. The question will then be: 'Find the domain and range of the function $f(x)=\sqrt{1-x^2}$, without any digital help'

Answer 1

Method 1:

Domain. In the real numbers $\sqrt{w}$ exists if and only if $w \ge 0$, so the domain is all real $x$ so that $1-x^2 \ge 0$ or $x^2 \le 1$ or $-\sqrt 1 \le x \le \sqrt 1$ or $-1 \le x \le 1$.

So domain is $[-1,1]$.

Range: $x$ may be anything so that $-1 \le x \le 1$. That means $x^2$ may be anything so that $0 \le x^2 \le 1$ That means $-x^2$ may be anything $-1 \le -x^2 \le 0$. Which means $1-x^2$ may be anything $0 \le 1-x^2 \le 1$ so $0 \le \sqrt{1- x^2} \le 1$.

So range is $[0,1]$.

Method 2:

The graph is $y = \sqrt{1-x^2} \ge 0$

So $y^2 = 1-x^2$

$y^2 + x^2 = 1$. That's a graph of a unit circle centered at $(0,0)$

So $y = \sqrt{1 -x^2}$ a half circle restricted to $y \ge 0$.

In this half circle, $x$ goes from $-1$ to $1$ so the domain is $[-1,1]$.

In the full circle $y$ goes from $-1$ to $1$ but we are restricting to $y\ge 0$ so the range is $[0,1]$.

Answer 2

I suppose you want this to be a real function. Then you have to ask yourself for which values this expression is real. Since $x^2$ is nonnegative, $x^2$ must not get bigger than $1$. The equation

$$x^2 =1 $$

has two solutions $1$ and $-1$. For any $|x|$ strictly greater than $1$, the function is not defined, for any $|x|$ smaller or equal $1$ it is defined. So you have found the domain to be $[-1,1]$.

For the range, we could try to find minima and maxima of this function. We differentiate once and find

$$f'(x) = -\frac{2x}{(1-x^2)^{1/2}} $$

Setting this equal to zero we find the root to be $x=0$. By differentiating once more and plugging in $x=0$ we see that there is a maximum at $x=0$ and it is $$f(0)=1$$

The minimum can then only be found at the boundary of the domain (we have only found one solution by looking at the roots of the derivative), i.e. $x=\pm-1$. There we have the value

$$f(\pm 1 )= 0$$

Therefore we have found the range to be $[0,1]$.

Answer 3

For real $x$ and real $f(x)$:

If $x\in dom (f)$ it is necessary that $1-x^2\geq 0$ (else $\sqrt {1-x^2}$ is not a real number). Therefore $$(i).\quad x\in dom (f)\implies 1-x^2\geq 0\implies x^2\leq 1 \implies |x|\leq 1.$$ And $1-x^2\geq 0$ is also sufficient for $x\in dom(f)$ because then the real number $\sqrt {1-x^2}$ exists. Therefore $$(ii).\quad |x|\leq 1\implies x^2\leq 1\implies 1-x^2\geq 0\implies x\in dom (f).$$ From (i) and (ii) we have $dom(f)=[-1,1].$

Now if $x\in dom (f)$ then $|x|\leq 1$. We have $$|x|\leq 1\implies 0\leq 1-x^2\leq 1\implies 0\leq \sqrt {1-x^2}\leq 1\implies 0\leq f(x)\leq 1.$$ Therefore $$ (iii).\quad ran (f)\subset [0,1].$$

For any $y\in [0,1]$ we have $1-y^2\in dom (f)$ and $f(\sqrt {1-y^2})=y.$ Therefore $$(iv).\quad ran (f)\supset [0,1].$$ From (iii) and (iv) we have $ran(f)=[0,1].$

Answer 4

Assuming you only work with real numbers, then the radicand must be non-negative, i.e. $1-x^2 \ge 0 \Rightarrow x^2 \le 1 \Rightarrow -1 \le x \le 1$.

Since $0 \le x^2 \le 1$ ($x^2$ is non-negative) we get that $\sqrt{1-x^2}$ is in $[0,1]$.

Answer 5

One way to determine the domain and range of $f(x) = \sqrt{1 - x^2}$ is to graph the function. If $y = \sqrt{1 - x^2}$, then $y \geq 0$. Moreover, \begin{align*} y & = \sqrt{1 - x^2}\\ y^2 & = 1 - x^2\\ x^2 + y^2 & = 1 \end{align*} which is the equation of the unit circle, that is, the equation of the circle with center at the origin and radius $1$.

The restriction that $y \geq 0$ means that we obtain the upper semi-circle.

From the graph, we can see that $f(x) = \sqrt{1 - x^2}$ has domain $[-1, 1]$ and range $[0, 1]$.

If we did not know how to graph the function, we would start by observing that for $f(x) = \sqrt{1 - x^2}$ to be real-valued, $1 - x^2 \geq 0$. Solving the inequality yields \begin{align*} 1 - x^2 \geq 0\\ 1 \geq x^2\\ 1 \geq |x|\\ |x| \leq 1 \end{align*} which implies that $-1 \leq x \leq 1$, from which we obtain the domain $[-1, 1]$.

We further observe that if $x$ is in the domain, $1 - x^2 \leq 1$, with equality holding only at $0$. Moreover, the square of a number in the domain is at most $1$, so $1 - x^2 \geq 1 - 1 = 0$. Since a polynomial function is continuous, $1 - x^2$ assumes every value in the interval $[0, 1]$, so we have $0 \leq 1 - x^2 \leq 1$. Taking square roots yields $0 \leq \sqrt{1 - x^2} \leq 1$, so the range is $[0, 1]$.

That said, it helps to recognize that the graph of $f(x) = \sqrt{r^2 - x^2}$ is the upper half of a circle with radius $r$ and center at the origin.