Decibel physics help

Re: Decibal

Making the square root zero is not a problem, making the argument (what's inside it) negative is. You can calculate when that is by first looking when it will be precisely zero though.

As for the second question: the domain of the function needs to be given. For example, you can define the function on [1, 2] only. By natural domain we mean the largest domain on which defining the function makes sense, for example [0, [itex]\infty[/itex]) for sqrt(x) and [itex]x \neq 0[/itex] for 1/x.

 

Re: Decibal

No, you don't know that! The argument of a square root certainly can be 0, it just can't be negative.
That is x(5-x) must be greater than or equal to 0. Now a product of numbers will be larger than or equal to 0, if and only if both factors are positive or 0, [itex]x\ge 0[/itex] and [itex]5- x\ge 0[/itex], or both factors are negative, [itex]x\le 0[/itex] and [itex]5- x\ge 0[/itex].

A function consists of (a) a domain- the possible values of x and (b)a rule for finding the y value corresponding to each x value. Often we are given only the function ("rule") y= f(x). In that case, the presumed domain, the "natural domain" is the set of all x values for which the equation can be calculated. For example, if [itex]f(x)= \sqrt{x}[/itex], then the "natural domain" is the set of all non-negative real numbers. But we are also free to state the domain as a subset of that. For example, I can define the function F(x) by the rule [itex]f(x)= \sqrt{x}[/itex] with the domain restricted to "all x larger than 1". That is now a different function than f and has domain "all real numbers larger than 1". Or I could define g(x) by [itex]g(x)= \sqrt{x}[/itex] with domain "all positive integers" or G(x) by [itex]G(x)= \sqrt{x}[/itex] with domain "all positive rational numbers". Those are 4 different functions, with the same rule but different domains. The "natural domain" for the rule [itex]y= \sqrt{x}[/itex] is the set of all non-negative real numbers.

Why was this titled "decibal"?