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Based on this answer, I tried to find the condition to function $y = \dfrac{a x+ b}{c x +d}$ where $ad \neq 0$ and $a d - b c \neq 0$ increasing on $(-d/c, +\infty)$. I tried

Clear["Global`*"];
f[x_] = (a  x + b)/(c  x + d );
Reduce[ForAll[x, x > -d/c, Derivative[1][f][x] > 0], {a, b, c, d}]

I know that, the answer is $a d - bc > 0$. How can I get this result?

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  • $\begingroup$ One obstruction (it seems) is that Reduce is oriented toward solving for variables, not expressions. So it determines the solutions for three cases for a, namely a<0, a==0, and a>0. It's hard to put those cases back together, since those cases are solved for b, c, d in turn in a similar way resulting in their own sub-cases. I doubt this is what you want but it gives the desired result: Or @@ Select[! Eliminate[Derivative[1][f][x] == 0, {x}] /. {{Unequal -> Less}, {Unequal ->Greater}}, Reduce[Implies[#, ForAll[x, x != -d/c, Derivative[1][f][x] > 0]], Reals] &] $\endgroup$
    – Michael E2
    Commented Mar 30 at 15:24

2 Answers 2

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It is not easy to organize logical expressions to what one wnats to see. What I see is at best is

FullSimplify[ LogicalExpand[Reduce[ForAll[x, x > -d/c, f'[x] > 0], {a, b, c, d}]]]

$$(b|c|d)\in \mathbb{R}\land ((a=0\land ((c>0\land b<0)\lor (b>0\land c<0)))\lor (b c<a d\land a\neq 0))$$

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Resolve[ForAll[x,x>-d/c,f'[x]>0],Reals]

(c!=0&&b c-a d<0)||(d!=0&&b c-a d<0)

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