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Mathematics 22 Online
OpenStudy (trantom):

Is the average slope between two points on the graph y=sqrt(x) always 0? Because if you have 3 and 5 on the y axis it must also be -3 and -5, therefore the two slopes are opposites and cancel to zero... right?

OpenStudy (trantom):

to be clear, the bigger question is how to determine average slope between points which are on a non-function line. Although please specificly answer this one. Also, would you rewrite x=y^3+8 as y=cbrt(x-8) right?

OpenStudy (science0229):

No, because if the 2 random points on y=sqrt(x) are (a,b) and (c,d)... sqrt(a)=b --------> a=b^2 sqrt(c)=d --------> c=d^2 Slope between (a,b) and (c,d) is (d-b)/(c-a), or (d-b)/(d^2 - b^2). (Factor and cancel out (d-b), assuming that (a,b) and (c,d) aren't the same point.) (d-b)/((d-b)(d+b))=1/(d+b)

OpenStudy (studygurl14):

\[y = \sqrt{x}\] plug in 4 for x \[y = \sqrt{4} = 2\] (4, 2) plug in 9 for x \[y = \sqrt{9} = 3\] (9,3) Average of two points: (4 + 9)/2 , (2 + 3)/2 (13)/2 , (5)/2 Average of x coordinates : 6.5 Average of y coordinates: 2.5 Is that what you mean?

OpenStudy (science0229):

In conclusion, the slope between 2 points, (a.b), (c,d) that are on y=sqrt(x) is 1/(d+b), assuming that those 2 aren't the same points. Now, 1/(d+b) can never be zero, so the average slope can never be zero.

OpenStudy (science0229):

For you second question, yes. x=y^3 + 8 can be rewritten as y=cbrt(x-8) without any restrictions.

OpenStudy (trantom):

but science, if you pick x=4 then y is 2 AND -2 so you have two slopes to deal with

OpenStudy (science0229):

sqrt(4) is not -2 because square root of any positive real number gives you positive real number.

OpenStudy (science0229):

If you draw the graph, you'll notice that the range is restricted to [0,infinity)

OpenStudy (science0229):

Also, by the definition of a function, if 2 different x-values have the same y-value, that is not function.

OpenStudy (science0229):

So, no function can have (2,4) and (-2,4) as its points.

OpenStudy (science0229):

y=sqrt(x) is a function, so if x=4, y=2.

OpenStudy (trantom):

no it isn't xD -2*(-2)=4 and 2*2=4

OpenStudy (science0229):

So, -2*-2=2*2=4 ----> sqrt((-2)^2)=sqrt(4) ------> -2=2 Are you saying that?

OpenStudy (science0229):

Try to figure out what's wrong...

OpenStudy (trantom):

no lol sqrt((-2)^2) is sqrt(-2*(-2)) which is positive 4 so concluding -2=2 isn't correct. You would conclude with -2=-2 and 2=2

OpenStudy (science0229):

There is a real number property, which states that sqrt(a^2)=a. If you substitute -2 for a, what happens?

OpenStudy (trantom):

Are you denying that a negative number times a negative number is a positive number? Sorry, but thats fundemental. There are very few people in this world who would agree that the graph y=sqrt(x) only is in quadrant 1. And that isn't a real number property its a natural number property.

OpenStudy (science0229):

I'm not denying that. If a=2, sqrt(-2*-2)=-2 -------> sqrt(4)=-2 --------> 2=-2 What do you think is wrong here?

OpenStudy (trantom):

I hope this a friendly argument, math isn't something to get mad about :)

OpenStudy (science0229):

Don't worry. This won't get serious

OpenStudy (science0229):

Let's start over.

OpenStudy (trantom):

lets go back to the original question

OpenStudy (trantom):

I have checked with other people and so far i have gotten two answers. The average slope between two points on a non-function graph like y=sqrt(x) is undefined, or it's always 0. I think saying that it is zero makes more sense because the slope of the part in the 4th quadrant is the exact opposite of that in the first.

OpenStudy (science0229):

First thing first, y=sqrt(x) is a function.

OpenStudy (science0229):

It passes the vertical line test.

OpenStudy (science0229):

y=sqrt(x) is only defined at the 1st quadrant and the origin.

OpenStudy (trantom):

no its not. it fails the vertical line test.

OpenStudy (science0229):

the domain is restricted to [0,infinity), right?

OpenStudy (trantom):

yes, x must be equal to or greater than 0.

OpenStudy (trantom):

otherwise you have a square root of a negative number which is on the complex plane, not the real plane.

OpenStudy (science0229):

Great.

OpenStudy (science0229):

Now, when x=0, what's y?

OpenStudy (trantom):

sqrt(0)=0 of course

OpenStudy (trantom):

but the square root of, lets take 4 for example, is \[\pm2\]

OpenStudy (science0229):

Why?

OpenStudy (studygurl14):

because @science0229 (-2)^2 = 4 and (2)^2 = 4

OpenStudy (science0229):

I didn't ask that because I didn't know it.

OpenStudy (science0229):

Squar root of any positive real number ALWAYS give you positive answer. NEVER negative.

OpenStudy (science0229):

For example, sqrt(4) is not -2

OpenStudy (trantom):

that isn't the ONLY solution, but it is ONE of the solutions. Definition of "square root of x" is a number which when multiplied by itself gives the product of x. -2*(-2)=4 so one solution of sqrt(4) is -2.

OpenStudy (studygurl14):

It could be though, @science0229

OpenStudy (science0229):

Did you forget the restriction for square roots??????

OpenStudy (studygurl14):

in school, the answer is usually a positive number, unless they specifically ask for the negative root as well

OpenStudy (science0229):

That is specifically only for solving equations.

OpenStudy (trantom):

if it can be used in an equation, its true xD

OpenStudy (science0229):

Solving equations and taking a square root of a number is not the same thing...

OpenStudy (trantom):

If -2 wasn't a value for the square root of 4, than it looks like you have to correct just about every single math book ever written. Good luck with that.

OpenStudy (science0229):

No.

OpenStudy (science0229):

-2 is a solution of the equation x^2=4 -2 is not the answer for sqrt(4)

OpenStudy (science0229):

the reason that -2 is a solution of the equation x^2=4 is that negative number squared is always positive. the reason that -2 is not sqrt(4) is that square root of a positive number can NEVER be a negative number.

OpenStudy (trantom):

you mean the square root of a negative can never be a positive number. that would be imaginary.

OpenStudy (science0229):

No. Sqrt(4) is 2, which is positive number. It is not -2 because it is a negative number.

OpenStudy (science0229):

All of us are us are using the fact that sqrt(a^2)=a

OpenStudy (trantom):

You're repeating the same thing over and over, whats your PROOF. Find a definition that actually says negative numbers can't be solutions to the square roots of postives

OpenStudy (science0229):

But, there is a restriction that a>=0

OpenStudy (trantom):

where does anyone say that.

OpenStudy (trantom):

just look up "definition of square root" on google. It even says at the bottom "7 is A square root of 49" implying there are MORE THAN ONE. not 7 is THE square root of 49, 7 is A square root of 49.

OpenStudy (science0229):

There is a thing called square root, and there is a thing called principal and negative square root.

OpenStudy (science0229):

Square root always gives you POSITIVE answer.

OpenStudy (science0229):

Principal root always gives you POSITIVE AND NEGATIVE answers.

OpenStudy (science0229):

Negative root always gives you NEGATIVE answer.

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