Finding an elementary function Can someone please help me find any elementary function that satisfies


*

*$f(0) = 6$

*$f(1) = f(-1) = 4$

*$f(2) = f(-2) = 1$?
I have been trying for nearly an hour, but I still can't figure it out.  Only the points listed above matter. Nothing else matters (I don't care what $f(1.5)$ or $f(3)$ or $f(-100)$ are). 
The symmetry motivated me to try and use absolute value, but it didn't get me anywhere since there is nonconstant slope.
 A: let's assume the function be,
$$f(x)=ax^4+bx^2+c$$
now,we get,
$$f(0)=a.0^4+b.0^2+c=6........(1)$$
$$f(\pm 1)=a+b+c=4.....(2)$$
$$f(\pm 2)=16a+4b+c=1.....(3)$$
solving these equations we get,$a=\dfrac{1}{4},b=-\dfrac{9}{4},c=6$
Hence,
$$f(x)=\dfrac{1}{4}x^4-\dfrac{9}{4}x^2+6$$
A: Many functions work.  Here's one way to get a fairly simple one:
We can fit a quadratic through the values at $0,1,2$.  That leads us to $6-\frac 12(x^2+3x)$.  Now we can use the absolute value to extend this symmetrically to the negative values, arriving at $$f(x)=6-\frac {x^2}2-\frac {3|x|}2$$
Alternatively, we could fit a quartic through the five given points and thereby avoid the use of the absolute value, though at the cost of increasing the degree.  I think the quadratic (with the absolute value) is the more elementary solution.
A: Note that if given a function $h(x)$ where $h(-x)=x$, if we can find a function $g(x)$ such that $g(h(0))=6$, $g(h(1))=4$, $g(h(2))=1$, then we can take $f(x) = g(h(x))$.  Rakibul Islam Prince seems to have taken $h(x) = x^2$ and $g(x)=\frac14 x^2-\frac94x+6$. lulu seems to have taken $h(x) = |x|$ and $g(x) = 6-\frac12x^2-\frac32x$ (note that $|x|^2$ is the same thing as $x^2$), but their answer can also be derived from $h(x)=x^2$ and $g(x)=6-\frac12x-\frac32\sqrt{|x|}$.
