It is true that t<span>he function f(x)=x+π is a polynomial. Pi is considered to be a constant of about 3.14. </span>
The first one (2-x^3) is the correct one
Cross multiplication
or multiplying by a number which goes in both rates
so, this is a quadratic equation, meaning two solutions, and we have a factored form of it, meaning you can get the solutions by simply zeroing out the f(x).
![\bf \stackrel{f(x)}{0}=-(x-3)(x+11)\implies 0=(x-3)(x+11)\implies x= \begin{cases} 3\\ -11 \end{cases} \\\\\\ \boxed{-11}\stackrel{\textit{\large 7 units}}{\rule[0.35em]{10em}{0.25pt}}-4\stackrel{\textit{\large 7 units}}{\rule[0.35em]{10em}{0.25pt}}\boxed{3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bf%28x%29%7D%7B0%7D%3D-%28x-3%29%28x%2B11%29%5Cimplies%200%3D%28x-3%29%28x%2B11%29%5Cimplies%20x%3D%20%5Cbegin%7Bcases%7D%203%5C%5C%20-11%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cboxed%7B-11%7D%5Cstackrel%7B%5Ctextit%7B%5Clarge%207%20units%7D%7D%7B%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D%7D-4%5Cstackrel%7B%5Ctextit%7B%5Clarge%207%20units%7D%7D%7B%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D%7D%5Cboxed%7B3%7D)
so the zeros/solutions are at x = 3 and x = -11, now, bearing in mind the vertex will be half-way between those two, checking the number line, that midpoint will be at x = -4, so the vertex is right there, well, what's f(x) when x = -4?
![\bf f(-4)=-(-4-3)(-4+11)\implies f(-4)=7(7)\implies f(-4)=49 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{vertex}{(-4~~,~~49)}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20f%28-4%29%3D-%28-4-3%29%28-4%2B11%29%5Cimplies%20f%28-4%29%3D7%287%29%5Cimplies%20f%28-4%29%3D49%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7Bvertex%7D%7B%28-4~~%2C~~49%29%7D~%5Chfill)
We assume that the function is as follows:

Then, we have that:

Thus, if we have that:

Therefore, the value for the function is f^-1(2) = -1/2 or