0
$\begingroup$

Find the domain and range of: $$y=\sqrt {x^2-2x-8}$$

My Attempt: For domain, $y$ is defined iff $$x^2-2x-8\geq 0$$ $$(x-4)(x+2)\geq 0$$ $$x\in (-\infty, -2]\cup [4, \infty)$$

For Range, $$y=\sqrt {x^2-2x-8}$$ $$y=\sqrt {(x-1)^2 - 9}$$ $$y^2=(x-1)^2-9$$ $$y^2+9=(x-1)^2$$

$\endgroup$
  • $\begingroup$ Shouldn't the range be y =>0 as this is in the real plane? $\endgroup$ – Mohammad Zuhair Khan Dec 26 '17 at 1:54
  • $\begingroup$ Technically I am unsure what are we supposed to do. Anyway what you say is true enough. $\endgroup$ – Mohammad Zuhair Khan Dec 26 '17 at 1:57
  • $\begingroup$ What is the question here? $\endgroup$ – Dave Dec 26 '17 at 1:58
  • $\begingroup$ @Dave, what's the range of the function? $\endgroup$ – pi-π Dec 26 '17 at 2:00
  • $\begingroup$ You've found the domain. One way to find the range is to find the minimum of the function on its domain (call this point $a$). Then, since the function is clearly unbounded on this domain, the range will be $[f(a),\infty)$, since the range is certainly a subset of $[0,\infty)$. $\endgroup$ – Dave Dec 26 '17 at 2:03
1
$\begingroup$

When you squared $y$ you introduced spurious solutions. Your work on the domain shows $y$ can go to $0$ and when $x$ gets large in either direction $y$ gets large and positive. What does that tell you about the range?

$\endgroup$
  • $\begingroup$ @ Ross Millikan, No idea about that. Could you please explain? $\endgroup$ – pi-π Dec 26 '17 at 2:01
  • $\begingroup$ Your final equation has a solution $(6,-4)$ which the original equation does not. That might convince you that $-4$ is in the range, but the original requires $y \ge 0$ $\endgroup$ – Ross Millikan Dec 26 '17 at 2:07
1
$\begingroup$

Notice that the term inside of the square root goes from $0$ to $+\infty$ in either direction. Therefore, $y$ would also go from $0$ to $\infty$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.