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Problem:

Find the range of $f(x)=\sqrt{x-1}$

The problem I face is: is the range $[0,\infty),$ or is it $(-\infty,\infty)$? I had learnt that $\sqrt{x^2} = \pm x$. However, on the Net, I read that $\sqrt{x^2}=|x|$ ie the output of a square root function is positive. If the output of a square root is always positive then the Range of $\sqrt{x-1}$ is $[0, \infty)$ $$$$

Could someone help in clearing this doubt?

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    $\begingroup$ It is $[0,\infty)$ $\endgroup$ Commented Jun 15, 2015 at 13:47
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    $\begingroup$ $\sqrt{x^2}=|x|$ By definition $\sqrt{2}$ is the positive one and if you want to refer to the negative one it's $-\sqrt2$. $\endgroup$ Commented Jun 15, 2015 at 13:48
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    $\begingroup$ The $\pm$ sign is introduced when the squareroot is taken, not if an expression is already in a root , moreover notice that, $(-x)^2 = (x)^2 $ and the definition of $|x|$ is that is $-x$ when $x<0$ and $x$ when $x>0$ , notice how both cases are positive. So basically your fallacy is in the expression $\sqrt{x^2}=\pm x$ $\endgroup$
    – Someone
    Commented Jun 15, 2015 at 13:48
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    $\begingroup$ @BetterWorld what are you saying by $\sqrt{x^2}=\pm x$ $\implies$ $|x|=\pm x$ which is wrong. Well if you don't define for which $x$. $\endgroup$
    – Someone
    Commented Jun 15, 2015 at 14:03
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    $\begingroup$ Perhaps what you had learned previously was that $x^2=a^2$ has solutions $\pm a$? $\endgroup$
    – John Joy
    Commented Jun 15, 2015 at 20:43

5 Answers 5

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We think of $\sqrt{}$ as the positive square root. This is for convenience, but it is the consensus of all mathematicians. It is of course true that $(-x)^2 = (x)^2$, so you might even say something like $x$ and $-x$ are both square roots of $x^2$. In solving an equation like $(x+4)^2 = 25$, you can only "take the square root" of both sides if you remember that there can always be up to two square roots, and you would want to write $(x+4) = \pm\sqrt{25}$.

But mathematicians agree that $\sqrt{}$ should always refer to one number, as functions must do. Confusion would ensue if when you wrote $\sqrt{}$ and when I wrote $\sqrt{}$, we might be referring to two separate numbers. So we want $\sqrt{}$ to be a function. Your question emphasizes that $\sqrt{}$ is a function by referring to it as $f(x)$ and asking about its range. Functions must output a $single$ value for each input. This is why we define $\sqrt{ x^2} = |x|$. The range is $[0,\infty)$.

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    $\begingroup$ You meant "This is why we define $\sqrt{x^\color{red}{2}} = |x|$." $\endgroup$ Commented Jun 16, 2015 at 9:02
  • $\begingroup$ @N.F.Taussig yes, fixed. (you can make a (suggested) edit to the post yourself for things like this.) $\endgroup$
    – djechlin
    Commented Jun 16, 2015 at 15:45
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$\sqrt{z}$ is a function(Maps a point to a single point). So $\sqrt{z^2}$can only take a single value. By convention, $\sqrt{z}\ge0$. So,$\sqrt{z^2}=|z|$. So to answer your question, range will be $[0,\infty)$

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Here are a few fallacy i have observed in your answer.

  1. The expression $f(x)=\sqrt{x-1}$ ,first of all requires a domain, normally you'd assume that $x\geq1$ i.e., $f(x)\mid \forall x\in [1,\infty) \to f(x)\in [0,\infty)$

  2. Your expression, $\sqrt{x^2}=\pm x$ is invalid, it should be $\color{red}{\pm}\sqrt{x^2}=\pm x.$

  3. Normally, $\sqrt{x^2}=|x|$ which is defined as, $$\begin{cases}x& x\geq0 \\ -x & x<0 \end{cases}$$

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The definition of the square root includes that it is the positive square root:

$$\sqrt{x^2}=|x|$$ (which means that the solutions to $x^2=a$ are $x=\sqrt a$ and $x=-\sqrt a$).

As the modulus function $f(x)\ge0$ has a range $[0,\infty)$, this means that the range of $f(x)=\sqrt{x-1}$ is $[0,\infty)$ just like the modulus function, and the domain is $x\ge1$.

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  • $\begingroup$ editied; is it clearer now? $\endgroup$
    – danimal
    Commented Jun 15, 2015 at 13:55
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for square root function to be defined x-1>=0,or x>=1 there for the range of function is [0,+inf)

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