Let a,b be reals and f(x)=ax+b+9/x. Prove that there exists x0∈[1,9], such that |f(x0)|≥2.
What we are showing is that the function cannot be entirely constrained to a rectangle -2<f(x)<2 for all 1<x<9. Here's a graph showing it can almost be done.
https://www.desmos.com/calculator/lwosarmwqr
f'(x) = a - 9/x^2 which has a critical point at x=3/sqrt(a) (*)
f"(x) = 18/x^3 which is positive on [1,9] so the critical point is a local minimum: f(3/sqrt(a))=6sqrt(a)+b
We require this to stay in the rectangle, so
[1] 6sqrt(a)+b > -2
The endpoints must also be in the rectangle. f(1)=a+b+9 and f(9)=9a+b+1, so
[2] a+b+9 < 2
[3] 9a+b+1 < 2
https://www.desmos.com/calculator/oltax5b3km
Shows there is no point satisfying all three inequalities (Desmos doesn't take implicit formulas of a,b so I used a=x, b=y) where you can see the closest solution is a=1, b=-8.
(*) If a is negative there is no critical point. In this case, if f(1)-f(9) were under 4, we might be able to squeeze the graph in. However, f(1)-f(9)=-8a+8<4 implies a>1/2 which is a contradiction.
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Posted by Jer
on 2021-10-16 16:50:48 |
Calculus solution with graphs
