Question

find largest value for which x^12-x^9+x^4-x+1> 0

Answer 1

y = x¹² - x⁹ + x⁴ - x + 1 
              To find largest x  such that  y > 0.

Is there some typing mistake ??   OR,   find largest x for  y < 0 ??? 
Please check question, as  y  is always above 0.

y = x (x³ -1) (x⁸ + 1) + 1 

1) for  x < 0, first two factors are -ve. So y is positive.
2) for  0 < x < 1, the product of three terms is negative, but less than 1.
     So y is positive.
3) for  x > 1, the product is +ve. SO y is +ve.


see diagram.

Answer 2

${\huge {\overbrace {\underbrace {\red{question}}}}}$

x¹²-x⁹+x⁴-x+1 >0

${\huge {\overbrace {\underbrace {\red{solution}}}}}$

We need x : the polynomial takes strictly nonnegative values.

We break the domain of x and pair terms in inequalities such that the behavior of the function becomes obvious. Let's go for the domain -1<x<1 first.

For all such x, 1>|x|

x^4>| x^9|

x^12 is positive anyway.

Thus for |x|<1,the polynomial is clearly positive.

For |x|>1

x^12>x^9

x^4>x . Thus, for |x|>1,the function takes only positive values.

Thus, for real x, the polynomial is clearly positive.

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