Question

The largest interval for which x¹²-x⁹+x⁴-x+1 >0 is
1)1 <_x<infinity
2)-(infinity)<x<_-1
3)-1<_x<_1
4)-(infinity )<x<infinity

I am well aware of the answer that it's (4) but I haven't understood the question yet plz provide me the proper explanation along with detailed solution ​

Answer 1

Answer:

(-∞,∞)

Step-by-step explanation:

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 means, x € (-∞ , ∞).

Therefore the largest interval for which f(x) > 0 is (-∞,∞).

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