The largest interval for which x^12-x^9+x^4-x+1 0 is
Answers
Largest interval just means the 'domain of the function'. Here, domain(subset) for which f(x) > 0.
=> x¹² - x⁹ + x⁴ - x + 1 > 0
=> x¹² + x⁴ - x⁹ - x + 1 > 0
=> x⁴(x⁸ + 1) - x(x⁸ + 1) + 1 > 0
Notice that for every real value of x, x⁴(x⁸ + 1) will be positive.
For any positive value( ≥ 1) of x:
x⁴(x⁸ + 1) is always greater than x(x⁸ + 1)
Hence, x⁴(x⁸ + 1) - x(x⁸ + 1) is positive number(> 0
Thus, x⁴(x⁸ + 1) - x(x⁸ + 1) + 1 is greater than 0.
For any negative value of x:
x(x⁸ + 1) is -x(x⁸ + 1), hence,
x⁴(x⁸ + 1) - (-x)(x⁸ + 1) = x⁴(x⁸ + 1) + x(x⁸ + 1) which will always be positive.
Hence, x⁴(x⁸ + 1) - x(x⁸ + 1) is positive number
Thus, x⁴(x⁸ + 1) - x(x⁸ + 1) + 1 is greater than 0.
For any value b/w 1 and 0: xⁿ < x [n is even]
Therefore, x⁴ < x → x⁴(x⁸ + 1) < x(x⁸ + 1)
=> x⁴(x⁸ + 1) - x(x⁸ + 1) is -ve
=> x¹² - x⁹ + x⁴ - x is -ve
=> x(x - 1)(x² + x + 1)(x⁸ + 1)
Terms other x - 1 will always give a +ve value for x(0<x<1). x - 1 is negative and greater than -1. x(decimal no.) multiplied by any number less than 10 is always less than 1.
Thus, x¹² - x⁹ + x⁴ - x > - 1
=> x¹² - x⁹ + x⁴ - x + 1 > 0
Hence,
x⁴(x⁸ + 1) - x(x⁸ + 1) + 1 is greater than 0
In each case, x⁴(x⁸ + 1) - x(x⁸ + 1) + 1 is greater than 0, hence, we can say that this inequality doesn't depend on the value of x. It is true for every(any) real number. Thus, x belongs to R.
It is true for every(any) real number. Thus, x belongs to R.
It means, x € (-∞ , ∞).
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x € (-∞ , ∞)
hope this will help you