0
$\begingroup$

I am trying to construct a continuous differentiable function $f(x)$ that for $x_1$ and $x_2$ takes the value $0$ and have global minimum at these points, i.e. $f(x_1)=f(x_2)=0$ and $f'(x_1)=f'(x_2)=0$, where $x_2>x_1$.

No other local minimum can exist than those at $x_1$ and $x_2$.

So far, I have been able to construct a function $g(x)$ that in principle looks what I am looking for:

$$ g(x) = 2(e^{x-x_1}+e^{x_2-x})-\cos(2\pi x)-(1+2e). $$

A plot of this function for the constants $x_1=0$ and $x_2=1$ can be found here at WolframAlpha.

As can be seen in the plot, $g(x)$ fulfills the idea of only having two local and global minima, however these are not position at $x_1$ and $x_2$.

Q: Any suggestions on either how to modify $g(x)$ or how to construct a new function that fulfills the demands?

$\endgroup$
5
$\begingroup$

Try $f(x) = (x-x_1)^2(x-x_2)^2$.

Then $f(x) \geq 0$ for all $x$, and $f(x) = 0$ iff $x \in \{x_1, x_2\}$, and $f'(x_i) = 0$.

$\endgroup$
3
$\begingroup$

How about$(x-x_1)^2(x-x_2)^2?$

$\endgroup$
  • 2
    $\begingroup$ Bingo ${}{}{}{}{}{}$. $\endgroup$ – copper.hat Feb 13 '13 at 18:21
1
$\begingroup$

Assuming that $x_1 \neq x_2$ then a simple quartic will do the trick.

How about $y=(x-x_1)^2(x-x_2)^2$?

We see that $dy/dx = (x-x_1)(x-x_2)(2x-x_1-x_2)$ meaning that $x=x_1$ and $x=x_2$ are both stationary points. Moreover, $(d^2y/dx^2)(x_1) = 2(x_1-x_2)^2 > 0$ and $(d^2y/dx^2)(x_2) = 2(x_1-x_2)^2 > 0$. Hence $x=x_1$ and $x=x_2$ are both local minima.

What about the other turning point: $x=\frac{1}{2}(x_1+x_2)$? Well:

$$\left.\frac{d^2y}{dx^2}\right|_{x=\frac{1}{2}(x_1+x_2)} =-(x_1-x_2)^2 < 0 \, . $$

It follows that $x=x_1$ and $x=x_2$ are the only minima and they are on the same level, i.e. $y(x_1) = y(x_2)$. The only other extremum is a maximum.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.