A Common Mistake

Algebra Level 2

Which of the above is the graph of $f(x) = \sqrt{ x^2}$ ?

1 2 3 4

6 solutions

Trevor Arashiro
Oct 5, 2014

As we know $\sqrt {x^2}=|x|$ . Thus the graph of $y=|x|$ is the graph #3. $\square$

I have come here via the related facebook post. The question is different there, the exponent there is on the outside - making (1) be the right answer. It is very confusing - I hope, someone will check it.

Dávid Sebők - 5 years, 6 months ago

Same here. I pin pointed graph 1 as the answer on facebook, came here, and instantly clicked on the first option. But to my surprise, the answer was different. Then I realized the question here was different. They should definitely correct this.

Hamza Shariq - 5 years, 6 months ago

i made the same mistake here.. i checked only that and it confused me up

Gowrisankar Sreeram - 5 years, 6 months ago

It's just different question. Pay attention to detail. I've seen both and got them both correct, because I payed attention. The question shouldn't be changed because you saw it written differently somewhere else. That's asinine.

King De Jesús - 5 years, 6 months ago

@King De Jesús – Well... I have to disagree. Maybe I've misunderstood you or you have misunderstood us, but in my opinion this question should be the same as the related facebook question, especially, because they are connected via hyperlink. Edit: I have just checked and it has been corrected on facebook. :)

Dávid Sebők - 5 years, 6 months ago

Yeah, the problem on facebook was just different from this site

Vu Hoang - 5 years, 6 months ago

Prasuu Prasanth
Oct 7, 2014

The given expression is same as |x|. That's why it is always above the x-axis.

AGHHHHHHH, calculated it then went and looked at my denominators and not my answers and found -x to 0 to +x, darn that quick response LOL

Allan Marriott - 5 years, 6 months ago

I fell foul yet again of the fact the question was different between the Facebook post and the actual site. I didn't re-read before selecting my answer.

Chris Harden - 5 years, 6 months ago

Oli Hohman
Dec 11, 2015

The easiest way to answer this question is to analyze the domain of the function. Since the only potential restriction for even index radicals is that they can't contain negative numbers, and x^2 is always positive or equal to 0, this means that x can take on any real value for this function. This rules out options one and two, for sure. Since x^2 is positive or 0, and f(x)=sqrt(x^2), f(x) is also positive or 0. Therefore, the only picture that would make any sense is 3 (abs(x)).

Moderator note:

Good way of eliminating options.

contradicting your point "since x^2 is greater than 0 then f(x) has to be positive", take the example of x^2= 25, then f(x) can be equal to +/- 5 both. Therefore the answer should be y=+ or - X. If we are concerned only about the positive square root then y=|x| will hold good, but that isnt mentioned in the question

Shyam Kumar - 5 years, 6 months ago

I don't see any contradiction. f(x) = sqrt(x^2). If x = +/-5, then f(x) = sqrt((-5)^2) = 5, which is positive.

Oli Hohman - 5 years, 6 months ago

I agree! The correct answer should show an X because there are two solutions for every square root. The square root of 25 is +/-5.

Rachel Reddoch - 5 years, 6 months ago

The square root of 25, denoted as $\sqrt{25}$ is just 5.

If we wanted to solve $x^ 2 = 25$ , we say that the answer is $\pm \sqrt{25}$ , and not just $\sqrt{ 25 }$ .

Calvin Lin Staff - 5 years, 5 months ago

Atul Ramani
Oct 9, 2014

$f(x)=\sqrt{x^2} \Rightarrow f(x)=|x|.$ $\therefore$ the graph of $f(x)$ is same as that of $|x|.$ Hence 3 is the correct solution.

Niaz Ghumro
Oct 7, 2014

3 is the option because it is equal to y=abs(x) so it is in the first and second quadrant.

Betty BellaItalia
Apr 15, 2017

A Common Mistake

6 solutions

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