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.
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Verified answer
Y = x¹² - x⁹ + x⁴ - x + 1To 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.
x¹²-x⁹+x⁴-x+1 >0
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.
