Suppose f(x) is a polynomial of degree four,h | vedclass

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(C) Since $f(x)$ is a polynomial of degree $4$,its derivative $f'(x)$ is a polynomial of degree $3$.Given that the critical points are $-1, 0, 1$,we h

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$4$

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(C) Since $f(x)$ is a polynomial of degree $4$,its derivative $f'(x)$ is a polynomial of degree $3$.Given that the critical points are $-1, 0, 1$,we have $f'(x) = k(x+1)(x)(x-1) = k(x^3 - x)$ for some constant $k 
eq 0$.Integrating $f'(x)$,we get $f(x) = k(\frac{x^4}{4} - \frac{x^2}{2}) + C$.We want to find $T = \{x \in \mathbb{R} \mid f(x) = f(0)\}$.Setting $f(x) = f(0)$,we have $k(\frac{x^4}{4} - \frac{x^2}{2}) + C = C$.This simplifies to $\frac{x^4}{4} - \frac{x^2}{2} = 0$,which implies $x^2(\frac{x^2}{4} - \frac{1}{2}) = 0$.Thus,$x^2 = 0$ or $x^2 = 2$.The elements of $T$ are $0, \sqrt{2}, -\sqrt{2}$.The sum of squares of these elements is $0^2 + (\sqrt{2})^2 + (-\sqrt{2})^2 = 0 + 2 + 2 = 4$.

Suppose $f(x)$ is a polynomial of degree four,having critical points at $-1, 0, 1$. If $T = \{x \in \mathbb{R} \mid f(x) = f(0)\}$,then the sum of squares of all the elements of $T$ is

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