<s>[INST] Maximum and Minimum value of an implicit function For the real value of $x$, $f\left( x \right)$ satisfies $f{\left( x \right)^3} - f{\left( x \right)^2} - {x^2}f\left( x \right) + {x^2} = 0$. When the maximum value of $f(x)$ is $1$ and the minimum value of $f(x)$ is $0$, what is the value of $f\left( { - \frac{4}{3}} \right) + f\left( 0 \right) + f\left( {\frac{1}{2}} \right) = \_\_\_\_\_$
My approach is as follow, as it is an implicit function we need to find the roots is $f(x)$.
We end up getting $(f(x)-1)(f(x)+x)(f(x)-x)=0$, so we end up getting three function viz.
$f(x)=1$; $f(x)=-x$ & $f(x)=x$ but how do we proceed further
 [/INST] If $f(x)^3-f(x)^2-x^2f(x)+x^2=0$ for every $x$, you can input the needeed values $x_0=0,-\frac{4}{3},\frac{1}{2}$ and solve the polynomial for $f(x_0)$. In general, this may give you 3 solutions for each $x_0$ (so 27 possible solutions for the problem), but maybe those 27 are the same.
Edit: The conditions $\max(f) = 1$ and $\min(f) = 0$ make two of the roots for $x_0 = -\frac{4}{3}$ impossible as well as one for $x_0 = \frac{1}{2}$. But still this don't have a unique answer. I found: $$ f(0)+f\left(\frac{1}{2}\right) + f\left(-\frac{4}{3}\right) \in \left\{ \frac{3}{2}, \frac{5}{2}, 3 \right\}$$
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