# 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

• Do you know if the domain is limited? Satisfactory values lie on the lines $y=1$, $y=x$, and $y=-x$. Constant functions have no distinct values for max and min, so $y=1$ isn't allowed. The remaining curves have no finite extrema. Mar 2, 2022 at 22:48

$$y=f(x). y^3-y^2-x^2y+x^2=0$$. Max $$y$$ is $$1$$. Min $$y$$ is $$0$$. Find $$f(-4/3)+f(0)+f(1/2)$$

$$g(x,y)= (y-x)(y+x)(y-1)=0$$.

The constraint requires only certain $$y$$ values for a given $$x$$.

$$x=0\implies y\in\{0,1\}$$

$$x=1/2\implies y^3-y^2-y/4+1/4=0.\implies (2y-1)(2y+1)(y-1)=0\implies y\in\{1/2,-1/2,1\}$$

$$x=-4/3\implies y^3-y^2-16y/9+16/9=0\implies (9y^2-16)(y-1)=(3y-4)(3y+4)(y-1)=0$$

$$\implies y \in \{4/3, -4/3,1\}$$

I'm getting only 18 possibilities, but those should be pared down. Not sure how to use the min and max values. $$f(x)\equiv1$$ doesn't allow a minimum value of $$0$$, so the $$1$$'s can be ignored.

Consistent definition of $$f(x)$$ requires consistently using $$f(x)\equiv x$$ or $$f(x)\equiv -x$$. So, combinations are reduced still further to $$\{-5/6, 5/6\}$$

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: $$f(0)+f\left(\frac{1}{2}\right) + f\left(-\frac{4}{3}\right) \in \left\{ \frac{3}{2}, \frac{5}{2}, 3 \right\}$$