Here is my solution. Let the given expression equal to y. y = (x^2 - ab)/(2x - a - b), calling alpha = a and beta = b it can be turned into a quadratic in x. that is x^2 - 2yx - (a + b)y - ab = 0 Notice that, for any value of y in the range, the above quadratic equation must have two real roots. Using the discriminant of the quadratic, (-2y)^2 - 4((a+b)y - ab) >= 0 that is y^2 - (a+b)y + ab>=0 Let's solve the inequality by factoring the left side. Observe that the given expression is symmetric in a and b. So, without loss of generality, let's assume a > b. (y-a)(y-b)>=0 It solves to give. y >= a or y
After finding f(a) = a, and f(b) = b, I found the derivative f'(x) = 2(x-a)(x-b) / (2x-a-b)^2. We can see that a and b are critical points, and then use the sign of f'(x) along with the fact that f is differentiable everywhere except for the asymptote at the midway point between a and b to deduce that f(x) never takes on values strictly between a and b.
A much easier solution: m = (alpha + beta)/2, we write alpha = m + d, beta = m - d x = m + t The original expression becomes E = [(m + t)^2 - (m^2 - d^2)] / (2t) = m + 1/2(d^2/t + t) >= m + d or
assume a < b WLOG, and let f(x) be the objective function. Note, there is a pole of f at x = (a + b)/2 , and so f -> +- inf at x -> (a + b)/2. Note that f(a) = a, f(b) = b, and f’(x) = [(2x-a-b)(2x) - 2(x^2 - ab)]/[(2x-a-b)^2], and thus has zeros at x^2 - (a+b)x + ab = 0 ==> a,b are the only stationary points of f. As x->+-inf, f ~ +-x/2 -> +-inf. Thus, supposing a,b aren’t inflections, by continuity of f and f’ and using b > a, we must have f(a) be a local maximum and f(b) a local minimum, which means f = f(b) = b over R, which gives the desired result. No inflections: this is immediate by the fact that there is a pole as x-> (a+b)/2, which means f->-inf as x-> (a+b)/2-, and f->inf as x-> (a+b)/2+ , and we cannot have any inflections connecting the stationary points at a,b due to this discontinuity, which is all we need to show. (For completeness, we also consider a = b ==> f(x) = (x^2 - a^2)/(2x-2a) = (x+a)/2 for all x =/= a, and hence cannot equal a, so we are fine. The geometric interpretation of this is that we bring b->a, and so our two stationary points tend towards the central discontinuity at their midpoint, which is at (a+b)/2 = a at the limit. What a great problem!)
Yeah, I thought this one was pretty cool too! I didn’t think of doing it by taking a derivative and doing some analysis, but I’m glad to see a couple of viewers leave comments with that approach!
Hi, thanks for doing question, essentially in the textbook it was how you ended up doing it, let function = y, cross multiply, end up with a quadratic in x, for solutions the discriminant >0, solve, end up with another quadratic in y, do same thing, show cannot lie between alpha and beta.
I did it basically the same way but as an explicit inverse function which cannot lie between alpha or beta. Although, at the end, I did it as an ineqality for the values of x rather than trying to place a value between alpha and beta.
The domain of a function is the set of all possible values of x, in this case any real number except (\alpha+\beta)/2 which would make the denominator zero. The range of a function is the set of all possible values that the function can attain, so in this case it is any real number not between \alpha and \beta as the question asked for.
I did it in a similar way, I just didn't call the expression "gamma" but left it as it is and assumed that this expression lies between alpha and beta. Then I transformed this inequality and shifted or added/subtracted some terms, which resulted in a contradiction.
The function f(x) has a pole at x= (a+b)/2 .Take a>b (a a and on the other side < b .
Here is my solution.
Let the given expression equal to y.
y = (x^2 - ab)/(2x - a - b), calling alpha = a and beta = b
it can be turned into a quadratic in x. that is
x^2 - 2yx - (a + b)y - ab = 0
Notice that, for any value of y in the range, the above quadratic equation must have two real roots. Using the discriminant of the quadratic,
(-2y)^2 - 4((a+b)y - ab) >= 0 that is
y^2 - (a+b)y + ab>=0
Let's solve the inequality by factoring the left side. Observe that the given expression is symmetric in a and b. So, without loss of generality, let's assume a > b.
(y-a)(y-b)>=0
It solves to give.
y >= a or y
After finding f(a) = a, and f(b) = b, I found the derivative f'(x) = 2(x-a)(x-b) / (2x-a-b)^2. We can see that a and b are critical points, and then use the sign of f'(x) along with the fact that f is differentiable everywhere except for the asymptote at the midway point between a and b to deduce that f(x) never takes on values strictly between a and b.
Interesting way of approaching it, didn’t think of doing it this way!
A much easier solution:
m = (alpha + beta)/2, we write alpha = m + d, beta = m - d
x = m + t
The original expression becomes
E = [(m + t)^2 - (m^2 - d^2)] / (2t) = m + 1/2(d^2/t + t) >= m + d or
assume a < b WLOG, and let f(x) be the objective function. Note, there is a pole of f at x = (a + b)/2 , and so f -> +- inf at x -> (a + b)/2. Note that f(a) = a, f(b) = b, and f’(x) = [(2x-a-b)(2x) - 2(x^2 - ab)]/[(2x-a-b)^2], and thus has zeros at x^2 - (a+b)x + ab = 0 ==> a,b are the only stationary points of f.
As x->+-inf, f ~ +-x/2 -> +-inf. Thus, supposing a,b aren’t inflections, by continuity of f and f’ and using b > a, we must have f(a) be a local maximum and f(b) a local minimum, which means f = f(b) = b over R, which gives the desired result.
No inflections: this is immediate by the fact that there is a pole as x-> (a+b)/2, which means f->-inf as x-> (a+b)/2-, and f->inf as x-> (a+b)/2+ , and we cannot have any inflections connecting the stationary points at a,b due to this discontinuity, which is all we need to show.
(For completeness, we also consider a = b ==> f(x) = (x^2 - a^2)/(2x-2a) = (x+a)/2 for all x =/= a, and hence cannot equal a, so we are fine. The geometric interpretation of this is that we bring b->a, and so our two stationary points tend towards the central discontinuity at their midpoint, which is at (a+b)/2 = a at the limit. What a great problem!)
Yeah, I thought this one was pretty cool too! I didn’t think of doing it by taking a derivative and doing some analysis, but I’m glad to see a couple of viewers leave comments with that approach!
Hi, thanks for doing question, essentially in the textbook it was how you ended up doing it, let function = y, cross multiply, end up with a quadratic in x, for solutions the discriminant >0, solve, end up with another quadratic in y, do same thing, show cannot lie between alpha and beta.
Thanks for the submission again!
I did it basically the same way but as an explicit inverse function which cannot lie between alpha or beta. Although, at the end, I did it as an ineqality for the values of x rather than trying to place a value between alpha and beta.
Can you explain the domain and range if composite functions
The domain of a function is the set of all possible values of x, in this case any real number except (\alpha+\beta)/2 which would make the denominator zero. The range of a function is the set of all possible values that the function can attain, so in this case it is any real number not between \alpha and \beta as the question asked for.
I did it in a similar way, I just didn't call the expression "gamma" but left it as it is and assumed that this expression lies between alpha and beta. Then I transformed this inequality and shifted or added/subtracted some terms, which resulted in a contradiction.
That’s a cool way of doing it too! I might try it like that some time.
wow well done
Thanks!