Finding the Range of a Function

We determine the range of the function f(θ) = (1 - tanθ)/(1 + cotθ). That is, the set of possible outputs, given real values of θ for which f(θ) is well-defined.
00:00 Domain
02:08 Substitution
03:34 Discriminant
05:34 Solving the inequality
06:41 Checking carefully: x = 0
08:03 Checking carefully: x = -1

[@user-cd9dd1mx4n]

Impressive! Your decision to create videos spanning diverse subjects like number theory, geometry, trigonometry, integration, etc, is commendable. It reflects a wise approach that encourages exploration across various areas of maths, rather than confining oneself to a single area of expertise. I appreciate that. For this problem, I tried to write f(θ) as f(θ)=[cos(θ)(cos(θ)-sin(θ))]/[cos(θ)(cos(θ)+sin(θ))], then somewhere in my solution, I failed to continue my attempt. Your video is truly nice, and I found it to be exceptionally well-done. Thanks for it.

[@DrBarker]

Thank you!

[@ianfowler9340]

As you did, I was able to find both the domain first (i.e. vertical asymptotes) and then the range before starting the video. Then graphed f(x) to check. I saw something very interesting and unexpected - for me anyway. If you look at the LOWER branches of f(x), they all have vertical asymptotes at regular intervals, but there are gaps between those asymptotes. At first glance, you might think that f(x) does not exist between those gaps. BUT when you look at the upper branches you will find that they fill in all of the gaps in the lower branches and with the same vertical asymptotes! And vice versa - the lower branches fill in the gaps bewteen the vertical asymptotes of the upper branches. Again with the exactly the same vertical asymptotes. And you can clearly see that f(x) does not exist between the minimums of the upper branches [3+2sqrt(2)] and the maximums of the lower branches [3-2sqrt(2)]. It's like the grand secret plan is all neatly buried and disguised inside the defining equation for f(x) and it's up to us to unravel it. That's what makes this so much fun. Thanks for such a nice problem. Addendum: I started by multiplying num. and denom. by tan(x) and that raised flags about 0, pi, 2pi ... I then took the right and left limits of f(x) as x---> 0 and found them both to be 0. At first I thought there might be a hole at (0,0). But then checked f(0), f(pi), f(2pi), ... and all was well.

[@mrphlip]

Solving this myself, I got as far as finding that expression y = x(1-x)/(x+1), like you have here, and then tried to figure out its range directly. If you graph that expression, you'll find that it has a pole at x=-1, and then to the left of that pole, it comes down from infinity, hits some local minimum, and then goes back up to infinity as x approaches negative infinity. Meanwhile, to the right of the pole, it has two roots at x=0 and x=1 (with x=0 being a hole), with a local maximum between them, and going down to negative infinity outside that. So, the range of the function is everything above that local minimum, and everything below that local maximum. All that's left is to find those values, and see if they overlap, or if there's a gap. In theory you could find those values in the usual way, by taking the derivative and finding the roots... but I took one look at that, and decided I CBA and just watched the video instead. Your solution with the quadratic formula is a lot more interesting as a solution, I think.

[@General12th]

Hi Dr. Barker! Fun!

[@Bayerwaldler]

Interesting! 👍My first instinct was: Oh, that looks boring

[@tioulioulatv9332]

الله يحفظك

[@mattikemppinen6750]

Hello and thank you for another great video! I saw the problem in a thumbnail the other day and independently worked it through before watching the video. I solved it using a different method and my range was (-inf,sqrt(2)/(3sqrt(2)+4)] U [sqrt(2)/(3sqrt(2)-4),inf). Checking numerically these endpoints seem to be the same as the ones in the video. I found it astonishing at first since they don't really look like something that should be equal to each other, but on a second glance they can easily be seen to be equal by equivalency by setting say sqrt(2)/(3sqrt(2)+4)=3-2sqrt(2) and multiplying through by the LHS denominator. This led me to wonder if there are any nifty results for checking if these kind of fractional expressions involving radicals could be written in a neater (non-fractional) form. My first guess is that it would have something to do with the divisibility among the terms of the numerators and denominators of these fractional expressions.

[@mattikemppinen6750]

Actually, twiddling with it a bit more you can see the factorization in this case quite easily when you reduce my expression a bit more by canceling the sqrt(2) in both numerator and denominator. In this case it just boils down to that (3-2sqrt2)(3+2sqrt2)=9-8=1. Still it would be cool to have some quick way to check for fairly simple expressions like this if it could be written in a neater form without actually multiplying it out and trying to figure it out. I feel like I might be missing something really obvious, but I guess that's enough Saturday Night Maths for now! =)

[@yurenchu]

The way to eliminate square roots in the denominator of a fraction, is to multiply both numerator and denominator by the so-called _conjugate_ of the denominator. The conjugate of (√a + √b) is (√a - √b) , and _vice versa_ [*]. The denominator then becomes "square root free" because (√a + √b)*(√a - √b) = (√a)² - (√b)² = a - b . In your particular case, the fraction is (√2)/[3√2 + 4] , and the square root in the denominator can be eliminated by multiplying both numerator and denominator by [3√2 - 4] . This results in (√2) / [3√2 + 4] = = ((√2)*[3√2 - 4]) / ([3√2 + 4]*[3√2 - 4]) = (√2)*[3√2 - 4] / ((3√2)² - 4²) = (√2)*[3√2 - 4] / (18 - 16) = (√2)*[3√2 - 4] / (2) = [3*2 - 4√2] / (2) = [3 - 2√2] I hope that helps. [*] Note: the term "the conjugate" is probably not entirely correct here, because (√b - √a) could also be considered a "conjugate" of (√a + √b) , and multiplying by (√b - √a) instead of by (√a - √b) would also work. However, I feel the precise terminology is not really important here, the point is that you'll know how to get rid of a square root sign in the denominator.

[@edmundwoolliams1240]

A tricky problem disguised as a friendly P3 A-level Maths exercise 😂

[@joeeeee8738]

I would have shown at the end an image of how the f(X) actually looks like to check

[@aravindhvijayanandan3010]

Wouldn't it be f(x) at 3:12 ??

[@ianfowler9340]

f(x) would still be [1-tan(x)] / [1+ cot(x)] - replace t with x - dummy variable ===> f( inverse tan(x) ) = (x - x^2) / (x + 1) g(x) = (x - x^2) /(x+1) would work with the rider that x = tan(t) Then g(x) = g( tan(t) ) = f(t) when x=tan(t)

[@yurenchu]

f(θ) = [1 - tan(θ)] / [1 + cot(θ)] = [1 - tan(θ)] / [1 + 1/tan(θ)] = tan(θ) * [1 - tan(θ)] / [tan(θ) + 1] = sin(θ) * [1 - tan(θ)] / [sin(θ) + cos(θ)] = sin(θ) * [cos(θ) - sin(θ)] / [sin(θ)cos(θ) + (cos(θ))²] = [sin(θ)cos(θ) - (sin(θ))²] / [sin(θ)cos(θ) + (cos(θ))²] = [sin(θ)cos(θ) - (1 - (cos(θ))²)] / [sin(θ)cos(θ) + (cos(θ))²] = [sin(θ)cos(θ) + (cos(θ))² - 1] / [sin(θ)cos(θ) + (cos(θ))²] = 1 - 1/[sin(θ)cos(θ) + (cos(θ))²] ... note: sin(2θ) = 2sin(θ)cos(θ) , and cos(2θ) = 2(cos(θ))² - 1 ==> (cos(θ))² = ½(cos(2θ)+1) ... = 1 - 1/[½sin(2θ) + ½(cos(2θ)+1)] = 1 - 2/[sin(2θ) + cos(2θ) + 1] ... note: sin(π/4) = cos(π/4) = ½√2 ... = 1 - 2/[ (√2)*( sin(2θ)cos(π/4) + cos(2θ)sin(π/4) ) + 1] = 1 - 2/[ (√2)*sin(2θ + π/4) + 1 ] The denominator of the second term is D(θ) = ((√2)*sin(2θ + π/4) + 1) ; the range of this denominator is [ 1-√2 , 1+√2 ] . Note that -2/(1-√2) = -2*(1+√2)/(1-2) = (2+2√2) , and -2/(1+√2) = -2*(1-√2)/(1-2) = (2-2√2) . Therefore, the range of the second term (-2/D(θ)) is (-ꝏ , 2-2√2] ∪ [2+2√2 , +ꝏ) , and the range of f(θ) = (1 - 2/D(θ)) is ( -ꝏ , 3-2√2 ] ∪ [ 3+2√2 , +ꝏ ) . EDIT: After watching the video, I see that I forgot to check for "invalid" values of θ , and the associated limit(s) of the function values. So although the answer I got is correct, my method wasn't rigourous.