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Finding a differentiable function $g:\mathbb{R}\rightarrow \mathbb{R}$ satisfying the following condiotions:

  • $\displaystyle g(0)=0, g(1)=1, g(-1)=-1;$

  • $\displaystyle g^\prime(1)=g^\prime(-1)=\frac{1}{2};$

  • $\displaystyle g^\prime(x)\geq\frac{1}{2} \quad \forall x\in [-1,1].$

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What have you already tried? What specifically is causing you difficulty? – user22805 Sep 25 '12 at 4:58

4 Answers 4

up vote 0 down vote accepted

You may be able to find values of $a$ and $b$ such that $y=a\arctan bx$ works.

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Thank you for your interesting hint. – blindman Sep 27 '12 at 2:45

Focus on constructing the derivative $h=g'$ first. It must meet certain requirements: $\int_0^1 h =1 = \int_{-1}^0 h$, etc., but all these can be fulfilled with a piecewise linear function with breaks at $0,-1,1$. Then integrate and get $g$.

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Thank you for your useful hint. – blindman Sep 27 '12 at 2:52

Following the useful hint of $\textbf{Gerry Myerson}$ we give the solution of my question. Let $g(x)=a\arctan(bx)$. We will choose $a,b\in \mathbb{R}$ satisfying the given conditions. We observe that $$ \begin{cases} g(0)=0&\\ g(1)=1&\\ g(-1)=-1& \end{cases} \Longleftrightarrow \quad a\arctan(b)=1. $$ and $$ g^{\prime}(x)=\frac{ab}{1+b^2x^2}\geq \frac{ab}{1+b^2}=g^{\prime}(1)=g^{\prime}(-1) \quad \forall x\in [-1,1], \forall a,b>0 $$ Therefore, to accomplish our goal it suffices to choose $a,b>0$ such that $$ \textbf{(I)}\quad \begin{cases} a\arctan(b)=1,&\\ \frac{ab}{1+b^2}=\frac{1}{2}.& \end{cases} $$ Let $$ h(b)=\displaystyle\frac{2b}{1+b^2}-\arctan(b), \quad b\in (0, +\infty) $$ Since $h(b)$ is continuous on $(0, +\infty)$, $h(1)=1-\frac{\pi}{4}>0$ and $$ \lim_{b\rightarrow +\infty}h(b)=-\frac{\pi}{2}<0 $$ there exists $b>0$ such that $h(b)=0$. Hence there exists $a,b >0$ satisfying $\textbf{(I)}$.

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Well done! Too bad it's just an existence result and doesn't give nice numbers for $a$ and $b$. – Gerry Myerson Sep 27 '12 at 7:00

Using the Fundamental Theorem you know that $\displaystyle\int_{-1}^1 g'(x) dx = 2.$ Plotting the known points of $g'$ together with the constraint shows that half of this area is accomodated by a rectangular box. The other half is easily obtained by constructing an isoceles triangle of area one to sit on top of it. You can check that the resulting $g$ has the remaining desired properties.

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Thank you for your useful hint. – blindman Sep 27 '12 at 2:53

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