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I came across the problem of finding range of the function, $$f(x) = \left(1-\sqrt{x}\right)^2$$

I proceed as follow: $$y = \left(1-\sqrt{x}\right)^2 $$ $$As, \left(1-\sqrt{x}\right)^2 \ge 0$$ $$So, y \ge 0$$ Hence, range of the given function is $\left[0,\infty\right).$

I know the answer is correct, but I am not sure about the process. Is this the correct way to do this ? If not, How to solve this problem ? I couldn't solve this by other methods.

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    $\begingroup$ An alternative to Kavi Rama Murthy's approach would be if you know or can prove that $(1 - \sqrt{x})^2$ is a continuous function that goes to $\infty$ as $x$ goes to $\infty.$ $\endgroup$ Apr 1, 2021 at 8:48

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You proved that the range is contained in $[0,\infty)$ not that it is equal to $[0,\infty)$. You have to take any $y$ in this interval and show that $y$ is actually in the range. For this you have to solve the equation $(1-\sqrt x)^{2}=y$. Can you do that?

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  • $\begingroup$ I couldn't solve that equation to find range. $\endgroup$ Apr 1, 2021 at 8:48
  • $\begingroup$ In case you wish to disagree, see the comment following the query that I just left $\endgroup$ Apr 1, 2021 at 8:49
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    $\begingroup$ $x=(1+\sqrt y)^{2}$ is a solution. @RohitJoshi $\endgroup$ Apr 1, 2021 at 8:50
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    $\begingroup$ @RohitJoshi Yes, but for this question it is enough to find one value of $x$ such that $f(x)=y$. $\endgroup$ Apr 1, 2021 at 8:58
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    $\begingroup$ @RohitJoshi Once you find $x$ such that $f(x)=y$ (for any $y\in [0,\infty)$) you are done. By definition of range that would prove that the range of $f$ is $[0,\infty)$ $\endgroup$ Apr 1, 2021 at 9:09
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$\sqrt x=1\pm \sqrt y\ge 0\implies \pm \sqrt y\ge -1$

Noting that $y\ge 0$, we have two cases:

Case 1: $\sqrt y\ge -1$
This is true for all $y\in [0,\infty)$
Case 2: $ -\sqrt y\ge -1\implies \sqrt y\le 1\implies 0\le y\le 1\implies y\in [0,1]$
Therefore the range is $[0,1]\cup [0,\infty)=[0,\infty)$

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Alternative approach to completing the problem.

To show:
$f(x) = (1 - \sqrt{x})^2$ is surjective on $[0,\infty)$ as $x$ goes from $1$ to $\infty$.

$\sqrt{x}$ is known to be a continuous strictly increasing function. Therefore, so is $f(x)$.

Further since $\sqrt{x} \to \infty$ as $x\to \infty, f(x)$ is unbounded, as $x \to \infty.$

Suppose there exists $r \in \Bbb{R^+}$ such that $r$ is not in the range of $f$. Since $f$ is unbounded, choose $s$ such that $f(s) > r.$

Consider the closed interval $[1,s]$. By the intermediate value theorem, $f$, being continuous, takes on every value between $f(1)$ and $f(s)$. Therefore, since $0 < r < f(s)$, there must be some value $t$ in $[1,s]$ such that $f(t) = r.$

This yields a contradiction. Therefore, it can not be the case that there exists any $r \in \Bbb{R^+}$ that is outside the range of $f$.

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  • $\begingroup$ Beautiful Proof. Sadly couldn't accept mutiple answers. But thank you ! $\endgroup$ Apr 1, 2021 at 9:15
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In fact, you proved nothing because you directly stated the answer

$$(1-\sqrt x)^2\ge0.$$

This is a true statement, but you did not verify if all non-negative values can be reached. A more complete discussion is

$$x\ge0\implies1-\sqrt x\in[-\infty,1)=(0,1]\cup(-\infty,0]\implies(1-\sqrt x)^2\ge0$$ and the inequalities are tight.

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    $\begingroup$ Any justification for this downvote ? $\endgroup$
    – user65203
    Apr 1, 2021 at 8:52
  • $\begingroup$ Do you disagree with the comment that I left, following the query? No idea who downvoted you.\ $\endgroup$ Apr 1, 2021 at 8:52
  • $\begingroup$ @user2661923: no, this is not enough. $\endgroup$
    – user65203
    Apr 1, 2021 at 8:54
  • $\begingroup$ Interesting, I thought that the Intermediate value theorem would kick in. $\endgroup$ Apr 1, 2021 at 8:57
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    $\begingroup$ @YvesDaoust But range of $1-\sqrt x$ is $1$ to $-\infty$. $\endgroup$
    – Eisenstein
    Apr 1, 2021 at 8:59

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