Let f:ℝ→ℝ be twice differentiable such that f(0)=2, f'(0)=-2, f(1)=1
Find for at-least how many c ∈ (0, 1)
f(c).f'(c)+f''(c)=0
is satisfied.
The simplest function that fits seems to be the quadratic f(x)=x^2-2x+2.
But then we require (c^2-2c+2)(2c-2)+2=0 which is a cubic with solutions around {-.9, 1.3, 2.4}. There are none on the interval (0,1). This implies the there are at-least zero c.
When I bumped up to a cubic there is more wiggle room. f(x)=ax^3+(1-a)x^2-2x+2 fits the qualifications for any a.
Playing with Desmos, I can find values of a with 2 solutions (if a=-0.5 then c=0.206 or c=0.791) or 1 solution (if a=0.5 then c=0.347) but I cannot find where there are zero values of c.
A question on notation then: Does f:ℝ→ℝ rule out quadratic equations?
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Posted by Jer
on 2019-06-25 09:17:38 |
Either solution or confused