What would be the correct way to construct a parabola that tangent to the function $\max \left\{ {x,0} \right\}$ and lies above it?
One of the thing that is difficult is that the max function is continuos but it seem that its derivative does not exist. My effort so far has been to be stucked at this step $\max \left\{ {x,0} \right\} = \frac{{x + 0 + \left| {x - 0} \right|}}{2}$.
The answer from the text book is that the parabola is the righ hand side of the inequality $\max \left\{ {x,0} \right\} \le \frac{1}{{4\left| {{x_0}} \right|}}{\left( {x + \left| {{x_0}} \right|} \right)^2}$
By inspection, I can observe that there are 2 points of contact.
Here $x_0$ is the touching point between the parabola and the function $\max \left\{ {x,0} \right\}$ and ${x_0} \ne 0$.
Please help me with this,
Thank you for your enthusiasm !
Since the parabola is tangent to the $x$ axis at some point $(x_0, 0)$ where $x_0 \lt 0$, then its equation is
$ y = a (x - x_0)^2$
The line $ y = x$ intersects this parabola at
$ x = a (x - x_0)^2 $
which gives the quadratic equation
$ a x^2 + x ( -2 a x_0 -1) + a x_0^2 = 0 $
If this equation is to have a single solution, then the discriminant has to be zero.
$ (-2 a x_0 - 1)^2 - 4 a^2 x_0^2 = 0 $
And this simplifies to
$ 4 a x_0 + 1 = 0 $
Therefore,
$ a = - \dfrac{1}{4 a x_0} $ Remember $x_0 \lt 0 $
This completes the specification of the parabola.
So our parabola is
$ y = - \dfrac{(x - x_0)^2}{4 a x_0 } = \dfrac{ (x + |x_0| )^2 }{ 4 a | x_0 | }$