catty2004 wrote:
Can someone please explain the signs in red above? this is not absolute value, why do we need to test these?
Actually you can transform it to an absolute value problem: \(1-x^2\geq{0}\) --> \(x^2\leq{1}\), since both parts of the inequality are non-negative then we can take square root: \(|x|\leq{1}\) --> \(-1\leq{x}\leq{1}\).
Now, other approach would be: \(1-x^2\geq{0}\) --> \(x^2-1\leq{0}\) --> \((x+1)(x-1)\leq{0}\) --> the roots are -1 and 1 --> "<" sign indicates that the solution lies between the roots, so \(-1\leq{x}\leq{1}\).
Solving inequalities:
x2-4x-94661.html#p731476 (
check this one first)
inequalities-trick-91482.htmldata-suff-inequalities-109078.htmlrange-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535everything-is-less-than-zero-108884.html?hilit=extreme#p868863Now, about x2suresh's approach: we have \((x+1)(x-1)\leq{0}\), so the product of two multiples is less than (or equal to) zero, which means that the multiples must have opposite signs. Then x2suresh checks the case A. when the first multiple (x+1) is negative and the second (x-1) is positive and the case B. when the first multiple (x+1) is positive and the second (x-1) is negative to get the range for which \((x+1)(x-1)\leq{0}\) holds true. Notice that, for this particular problem, we don't realy need to test case A, since it's not possible (x+1), the larger number, to be negative and (x-1), the smaller number to be positive. As for case B, it gives: \(x+1\geq{0}\) and \(x-1\leq{0}\) --> \(x1\geq{-1}\) and \(x\leq{1}\) --> \(-1\leq{x}\leq{1}\).
Hope it helps.