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?
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?
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)
\[y = \sqrt{x}\] 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?
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.
For you second question, yes. x=y^3 + 8 can be rewritten as y=cbrt(x-8) without any restrictions.
but science, if you pick x=4 then y is 2 AND -2 so you have two slopes to deal with
sqrt(4) is not -2 because square root of any positive real number gives you positive real number.
If you draw the graph, you'll notice that the range is restricted to [0,infinity)
Also, by the definition of a function, if 2 different x-values have the same y-value, that is not function.
So, no function can have (2,4) and (-2,4) as its points.
y=sqrt(x) is a function, so if x=4, y=2.
no it isn't xD -2*(-2)=4 and 2*2=4
So, -2*-2=2*2=4 ----> sqrt((-2)^2)=sqrt(4) ------> -2=2 Are you saying that?
Try to figure out what's wrong...
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
There is a real number property, which states that sqrt(a^2)=a. If you substitute -2 for a, what happens?
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.
I'm not denying that. If a=2, sqrt(-2*-2)=-2 -------> sqrt(4)=-2 --------> 2=-2 What do you think is wrong here?
I hope this a friendly argument, math isn't something to get mad about :)
Don't worry. This won't get serious
Let's start over.
lets go back to the original question
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.
First thing first, y=sqrt(x) is a function.
It passes the vertical line test.
y=sqrt(x) is only defined at the 1st quadrant and the origin.
no its not. it fails the vertical line test.
the domain is restricted to [0,infinity), right?
yes, x must be equal to or greater than 0.
otherwise you have a square root of a negative number which is on the complex plane, not the real plane.
Great.
Now, when x=0, what's y?
sqrt(0)=0 of course
but the square root of, lets take 4 for example, is \[\pm2\]
Why?
because @science0229 (-2)^2 = 4 and (2)^2 = 4
I didn't ask that because I didn't know it.
Squar root of any positive real number ALWAYS give you positive answer. NEVER negative.
For example, sqrt(4) is not -2
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.
It could be though, @science0229
Did you forget the restriction for square roots??????
in school, the answer is usually a positive number, unless they specifically ask for the negative root as well
That is specifically only for solving equations.
if it can be used in an equation, its true xD
Solving equations and taking a square root of a number is not the same thing...
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.
No.
-2 is a solution of the equation x^2=4 -2 is not the answer for sqrt(4)
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.
you mean the square root of a negative can never be a positive number. that would be imaginary.
No. Sqrt(4) is 2, which is positive number. It is not -2 because it is a negative number.
All of us are us are using the fact that sqrt(a^2)=a
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
But, there is a restriction that a>=0
where does anyone say that.
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.
There is a thing called square root, and there is a thing called principal and negative square root.
Square root always gives you POSITIVE answer.
Principal root always gives you POSITIVE AND NEGATIVE answers.
Negative root always gives you NEGATIVE answer.
Join our real-time social learning platform and learn together with your friends!