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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

Samar Imam Zaidi
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  • 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. – TurlocTheRed Mar 02 '22 at 22:48

2 Answers2

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$y=f(x). y^3-y^2-x^2y+x^2=0$. 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 $\{-5/6, 5/6\}$

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

Rafael H.
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