• anonymous
at in example 3, it says that interval can’t be the interval of validity because it contains x = 0 and we need to avoid that point for -inf
  • Stacey Warren - Expert
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  • schrodinger
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  • phi
he important idea is that the solution for y is a function of x: y= f(x) Now for a particular f(x), there may be x's that result in an undefined y value. For example if y= sqrt(x)/2 then x<0 must be excluded. In Example 3 y= (3+sqrt(9+12x^2-4x^3))/x^2 y will be undefined when x=0 because of the division by x^2. y will also be undefined if the expression in the sqrt is negative. The example found that the expression under the radical sign is ≥0 when x≤3.2... However, because we are dividing by x^2, we must exclude 0. So we divide the interval into two parts, excluding the zero: -inf < x < 0 and 0 < x < 3.2... The valid interval is the one that contains the initial value -1. So we choose -inf < x < 0 If you look at the previous examples, f(x) did not contain a division by x, so x=0 was ok
  • anonymous

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