@MasterThief1324 √(9) is the principal root of 9 which is the non-negative root. If you want to refer to both roots, you need to write ±√(9) . This confused me when I first learned it in school.
I really confuse at 4:23 of the video why does the equation suddenly become like this aren't we have the rule stated that -a = -b will be turn in to a = b but in this case if you look at it clearly at what sal is doing he turn -a = -b into -a = b Is it legal to do so ?
Essentially you have the negative square root of a negative number squared. like -sq.rt.((-1)^2). so you square the negative number Which is now positive) take the square root of it then multiply it by a negative. pemdas works every time.
Dear Mr. Khan you’ve been super explicit, indeed a master you are. We begin with half of a parabola and advance to the inverse which graphically is much more open. I now see the profile of a boat’s hull. But really what could One do with an inverse. Please suggest one of your other presentations pointing to a practical purpose for running out an inverse.
one perhaps dumb question but I'd like to ask if the inverse of a function really is changing the dependent variable to the independent I mean if you have say y = 2x + 1 x = 1/2y - 1/2 if you graph the f(x) you get the same as if you graph the f(y), it is only when you change the y to x and x to y in the second function (the inverse) that you get the correct graph. Does it only become the inverse function when the switching of the y and x is done? I mean if you solve for x with respect to y you will still obviously get the same graph. Thanks!
Sir at 3:29 the answer for positive Sq. root of (-3)^2 won't be -3 instead of +3? Because, the root would just cancel out with power. Also, instead of the root itself negative sign can't we just consider the negative root (solution) of the number we are taking the square root of?
@Hrnmhmm ah no, I thought that for a sec at first. See tho if x is < (or equal to) 1, then for ANY value of x that you input, it will give a result of (x - 1) < (or equal to) 0, so it's -ve anyway. Also, you have to consider that your taking the -ve root of 1, so one has to be -ve also...
I don't understand why √((-3)^2) = 3. If you rewrite it in terms of exponents it would be: ((-3)^2)^.5 = (-3)^1 = -3. In the example you provided, wouldn't √((-3)^2) = √9 simplify to (-3) = ±3 (respectively). Thus √((x-1)^2) would leave you with (x-1) where the difference is still negative.
you lost me at why x-1 must be negative. ...never mind...just realized that using 1 or 0 for x will make x-1 undefined so they wont even apply for the function to work...therefore the only real solutions have to be negative.
@MasterThief1324 √(9) is the principal root of 9 which is the non-negative root. If you want to refer to both roots, you need to write ±√(9) . This confused me when I first learned it in school.
I really confuse at 4:23 of the video why does the equation suddenly become like this
aren't we have the rule stated that -a = -b will be turn in to a = b but in this case if you look at it clearly at what sal is doing he turn
-a = -b into -a = b
Is it legal to do so ?
3:41 not 0, as for 0, there is only 0 (& also isn’t it the same for non-perfect squares like 1/4, which is 1/2^2 & -1/2^2)
Essentially you have the negative square root of a negative number squared. like -sq.rt.((-1)^2). so you square the negative number Which is now positive) take the square root of it then multiply it by a negative. pemdas works every time.
please do harder examples like: f(x)= x^5+x^4+3
Dear Mr. Khan you’ve been super explicit, indeed a master you are. We begin with half of a parabola and advance to the inverse which graphically is much more open. I now see the profile of a boat’s hull. But really what could One do with an inverse. Please suggest one of your other presentations pointing to a practical purpose for running out an inverse.
one perhaps dumb question but I'd like to ask if the inverse of a function really is changing the dependent variable to the independent
I mean if you have say
y = 2x + 1
x = 1/2y - 1/2
if you graph the f(x) you get the same as if you graph the f(y), it is only when you change the y to x and x to y in the second function (the inverse) that you get the correct graph.
Does it only become the inverse function when the switching of the y and x is done? I mean if you solve for x with respect to y you will still obviously get the same graph.
Thanks!
Daski69 yea i think the x and y have to interchange
Sir at 3:29 the answer for positive Sq. root of (-3)^2 won't be -3 instead of +3? Because, the root would just cancel out with power. Also, instead of the root itself negative sign can't we just consider the negative root (solution) of the number we are taking the square root of?
@khanacademy, is it -1 at the end? not +1? why did you add 1?
A(B12)+C(D)=*^
@Hrnmhmm ah no, I thought that for a sec at first. See tho if x is < (or equal to) 1, then for ANY value of x that you input, it will give a result of (x - 1) < (or equal to) 0, so it's -ve anyway. Also, you have to consider that your taking the -ve root of 1, so one has to be -ve also...
I don't understand why √((-3)^2) = 3. If you rewrite it in terms of exponents it would be: ((-3)^2)^.5 = (-3)^1 = -3. In the example you provided, wouldn't √((-3)^2) = √9 simplify to (-3) = ±3 (respectively). Thus √((x-1)^2) would leave you with (x-1) where the difference is still negative.
how exactly do you plot the graph?
Do you still have the same question?
@MasterThief1324 Then again by PMDAS, √((-3)^2) = √(9) = ±3. But even then, couldn't you just "use" the "negative" component?
How is it negative one? should'nt it be negative two postitive one?
you lost me at why x-1 must be negative. ...never mind...just realized that using 1 or 0 for x will make x-1 undefined so they wont even apply for the function to work...therefore the only real solutions have to be negative.