While finding the Domain of this expression $$\sqrt{(2x-8)(x-1)}$$

I got this:


$2x-8\geq0 \vee x-1\geq0$

$x\geq4 \vee x\geq 1$

So the Domain is: $x\in [1, +\infty)$

But the real domain is

$x\in (-\infty,1] \cup [4,+\infty)$

So it means that $x\leq1$

Can someone please explain this?

  • 1
    $\begingroup$ You only considered $x\geq4 \vee x\geq 1$ where both factors are positive; but the product is positive also if both factors are negative. $\endgroup$
    – Davide
    Mar 2 at 8:59
  • $\begingroup$ @Davide Pardon me, but still don't get it (⁠╥⁠﹏⁠╥⁠) $\endgroup$
    – IudMG
    Mar 2 at 9:05
  • $\begingroup$ $(2x-8)(x-1)\ge0$ implies that $x\le1\vee x\ge4$. $\endgroup$
    – CY Aries
    Mar 2 at 9:11
  • 1
    $\begingroup$ Davide means that the radicand $(2x-8)(x-1) = 2x^2 - 10x + 8$ as a whole should be positive, which is different than imposing $2x-8 \geq 0$ and $x-1 \geq 0$ separately. $\endgroup$
    – Abezhiko
    Mar 2 at 9:12

2 Answers 2



$2x-8\geq0 \vee x-1\geq0$

An error occurs here. In writing $\lor$, you are saying "or," but this is not the case for this. In general, $ab \ge 0$ only if both are positive, both are negative, or at least one is zero. Hence, you would instead write

$$\Big( 2x - 8 \ge 0 \text{ and } x-1 \ge 0 \Big) \text{ or } \Big( 2x - 8 \le 0 \text{ and } x-1 \le 0 \Big)$$

For instance, by writing "or," in your original expression, if $2x-8 > 0$ but $x-1 < 0$, then you get a negative under the radical -- not good!

In the first case, you can see that $x \ge 4$; in the second, $x \le 1$. Why? In the first, you derive

  • $x \ge 4$, and
  • $x \ge 1$

Which values $x$ satisfy both? Well, all $x \ge 4$. Likewise, for the other, we derive

  • $x \le 4$, and
  • $x \le 1$

The values that satisfy both are $x \le 1$. Hence, we conclude that the domain of the function is all $x$ such that $$ x \ge 4 \text{ or } x \le 1 $$

I suspect your original error comes from using the "property"

$$\sqrt{ab} = \sqrt a \sqrt b$$

hence leading to you claiming

$$\sqrt{(2x-8)(x-1)} = \sqrt{2x-8} \sqrt{x-1}$$

but this does not hold all of the time! In the first I gave with $a,b$, we require $a \ge 0$ and $b \ge 0$, because otherwise the square roots on the right side could have negatives in their radicals -- not good! Likewise, $2x-8$ and $x-1$ could both be negative, making the expressions on the right side above ill-defined, but their product could be positive, keeping the original function well-defined just fine.

  • 1
    $\begingroup$ You meant to say that $ab \geq 0$ if both $a$ and $b$ are positive, both $a$ and $b$ are negative, or at least one of the numbers $a$ and $b$ is equal to zero. $\endgroup$ Mar 2 at 13:16

enter image description here

you can draw a sign table to get $D_f$ Domain fo the function


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