Solve: $x^{4} - x^{3} + x^{2} - x + 1 = 0$

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Solution

$f (x) = x^{4} - x^{3} + x^{2} - x + 1$
$f (0) = 1$
$f^{'} (x) = 4 x^{3} - 3 x^{2} + 2 x - 1$
$f^{''} (x) = 12 x^{2} - 6 x + 2 = 2 (6 x^{2} - 3 x + 1)$
We can see that the second derivative doesn't go to zero for any x.
Thus, first derivative is an monotonically increasing function.
First derivative is zero at $0.6 < x < 0.7$
The function gives $f (0.6) > 0 a n d f (0.7) > 0$
So, $f (x) \neq 0$ for all x

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