For $f(x): y=\sqrt{x-1}$, I have deduced that the domain for $x$ cannot include negative values, for negatives values for $x$ in this equation will cause the total value inside the square root sign to be negative (Thus, $x\geq 0$). However, I have argued my teacher that it is possible to have a negative value for y, since the square root of any value can equal a positive or negative value. (Ex: If $x=5$, $y=\sqrt{x-1}$ is equal to $y=\sqrt{4}$ and therefore $y=\pm 2$). I also have explained that if we have another take on it and rearrange the equation to become $y^2=x-1$, $y$ can definitely be a negative value, for it is squared and will not cause the value of $x$ to become negative. My teacher still thought I was wrong, so it'd be much appreciated if you give your take on this.
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Hint: can a square root be negative? – Sean Roberson Mar 06 '19 at 05:41
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(No, not when talking about principal square roots, which is what we usually do.) – Sean Roberson Mar 06 '19 at 05:41
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2Possible duplicate of Square roots -- positive and negative also see this, this, and this. – Ross Millikan Mar 06 '19 at 06:02
1 Answers
This is just a terminology error. The definition of $\sqrt x$ is the non-negative number that when squared equals $x$, a number we will call $a$. Now, the solutions of $x=a^2$ are $a$ and $-a$. The square root of a number squared is equal to that number, but the converse, the statement when a number squared is equal to a second number, that first number is the square root of the second, is false. To make an analogy, all squares are rectangles, but not all rectangles are squares. $a=\sqrt x$ is always a solution of $x=a^2$, but a solution of $x=a^2$ is not necessarily the square root of $x$. The reason is simply to make $y=\sqrt x$ a function, and make it a lot easier to work with. The reason why squaring to get the other equation does not work is because the squaring property of equality can change the domain of a function, as it does here, so the original domain must carry over.
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