How to find the range of $$\sqrt {x^2-5x+4}$$ where $x$ is real.
What I've tried:
Let $\sqrt {x^2-5x+4}=y$, solving for real $x$, as $x$ is real discriminant must be $D\geq0$. Solving I get $y^2\geq\frac{-9}{4}$. Which I suppose implies all real . But on wolfram alpha it says $y\geq0$. What am I doing wrong.
Any help appreciated.
Thank you.
The mistake is a very common one. Namely, what you did was transforming:
$$\sqrt{x^2-5x+4} = y$$
to:
$$x^2-5x+4 = y^2$$
Now, indeed, any solution $(x,y)$ of the first equation will produce one of the second. But, $z^2 = y^2$ does not imply $z =y$. It only implies $z = \pm y$. So it will have to be determined which of the two cases applies.
Some of the solutions you found to the quadratic actually correspond to cases where $\sqrt{x^2-5x+4} = -y$. This is because, by definition, the square root is always positive. Therefore, if $y < 0$, we can't have that $\sqrt{x^2-5x+4} = y$ -- it must be the $-y$ case.
I hope that clears the air for you.
Note that the domain of the function is $(-\infty,1]\cup [4, \infty)$, since
$$ y=\sqrt{(x-1)(x-4)}, $$
which tells you that $y\geq 0$
Note: Compare with the function
$$y=\sqrt{x}.$$
