Question

at http://tutorial.math.lamar.edu/Classes/DE/Exact.aspx 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 14 years ago

Answer 1

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

Answer 2

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