The problem is to find an explicit function $f(x)$ that satisfies the following conditions:

  • $\lim_{x \to 2^{-}}f(x) =- \infty $
  • $\lim_{x \to 2^{+}}f(x) = \infty $
  • $\lim_{x \to - \infty }f(x) =0 $
  • $f(-2)=2$
  • $f(5)=1$
  • $f(0)=0$
  • $f'(x)>0$ if $x<-2 $ or $x>5$
  • $f'(x)<0$ if $-2<x<2$ or $2<x<5$
  • $f'(5)=0$
  • $f'(-2)=0$
  • $f''(x)>0$ if $ x <-3$ or $x>2$
  • $f''(x)<0$ if $ -3 < x < 2$

I am not sure if there exists a rational function $ f(x)=\frac{g(x)}{h(x)}$ where $g,h$ are polynomials that satisfy the conditions. I have an idea of the general shape and believe an equation of the form $f(x)= \frac{a}{x-2}+ b^x $ with $b>1$ will satisfy the conditions.

My attempt: $$f(x) = \frac{1.115760990}{x-2}+1.085667362^x$$ This satisfies some but not all the conditions, shown in this graph.

I am thinking perhaps starting from the second derivative conditions we can work backwards to $f(x)$.

  • 1
    $\begingroup$ Hmm..atleast I could see that it doesnot satisfy $f(0) =0,f(5)=1$ ! $\endgroup$
    – BAYMAX
    Commented Nov 28, 2017 at 2:48
  • $\begingroup$ Is it known that there is a simple solution ? Are there additional clues ? $\endgroup$
    – user65203
    Commented Nov 28, 2017 at 9:33
  • 1
    $\begingroup$ A rational function is indeed not possible. Because the asymptote on the left requires the degree of the denominator to exceed that of the numerator. But then $f(\infty)=0$, which is contradictory with the fact that $f(5)=1$ is the rightmost minimum. $\endgroup$
    – user65203
    Commented Nov 28, 2017 at 9:36


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