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How to solve ? k>k3 , if k is non negative integer
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Method 1:
Think of the number line and the main 4 regions: #a) \(x < -1\) #b) \(-1 < x < 0\) #c) \(0 < x < 1\) #d) \(x > 1\)
Try a number from each of these regions to check the inequality \(x > x^3\) For #a) x = -2: -2 > -8 --- satisfies For #b) x = -1/2: -1/2 > -1/8 --- does not satisfy For #c) x = 1/2: 1/2 > 1/8 --- satisfies For #d) x = 2: 2 > 8 --- does not satisfy
Thus, x can be in regions #a or #c => \(x < -1\) or \(0 < x < 1\)
Trying x = 2, we have: x(x + 1)(x - 1) = 6 > 0 Thus, we have the curve of x(x + 1)(x - 1) as shown below. Since we needed to solve for \(x(x + 1)(x - 1) < 0\), the required solutions (regions) which are negative, i.e. below the X-axis, are shown in yellow:
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