Square Root (x(5-x)): Natural Domain & Proving Algebraically

  • Thread starter imdapolak
  • Start date
  • Tags
    Physics
In summary: The natural domain is the domain that makes sense for the function, in this case all non-negative real numbers.
  • #1
imdapolak
10
0
1. What is natural Domain of Square root(x(5-x))
2.
3. I know x can not make the square root equal zero, but I am not sure how to prove this algebraically. Any help would be appreciated, also what's the difference between natural doman, and Domain? My book doesn't explain this at all.
 
Physics news on Phys.org
  • #2


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.
 
  • #3


imdapolak said:
1. What is natural Domain of Square root(x(5-x))



2.



3. I know x can not make the square root equal zero,

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].

but I am not sure how to prove this algebraically. Any help would be appreciated, also what's the difference between natural doman, and Domain? My book doesn't explain this at all.
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"?
 

FAQ: Square Root (x(5-x)): Natural Domain & Proving Algebraically

What is the natural domain of the function f(x) = √(x(5-x))?

The natural domain of a function is the set of all real numbers for which the function is defined. In this case, the function f(x) = √(x(5-x)) is defined for all values of x that make the expression inside the square root non-negative. Therefore, the natural domain of this function is x ≤ 5 and x ≥ 0.

How do I prove algebraically that the natural domain of f(x) = √(x(5-x)) is x ≤ 5 and x ≥ 0?

To prove algebraically that the natural domain of a function is x ≤ 5 and x ≥ 0, we need to show that the expression inside the square root cannot be negative. This can be done by using the discriminant (b²-4ac) of the quadratic equation x(5-x). If the discriminant is greater than or equal to 0, then the expression is non-negative and the function is defined for all real values of x. In this case, the discriminant is 25-4(x)(5-x) = 5x-x², which is equal to 5x -x². This is greater than or equal to 0 for x ≤ 5 and x ≥ 0, as required.

Is the function f(x) = √(x(5-x)) defined for negative values of x?

No, the function f(x) = √(x(5-x)) is not defined for negative values of x. This is because the expression inside the square root becomes negative when x is negative, which is not allowed since the square root of a negative number is undefined in the real number system.

Can I simplify the function f(x) = √(x(5-x))?

Yes, the function f(x) = √(x(5-x)) can be simplified. By expanding the expression inside the square root, we get f(x) = √(5x-x²). This can be further simplified to f(x) = √(x(5-x)) = √(x)√(5-x) since the square root of a product is equal to the product of the square roots.

What is the range of the function f(x) = √(x(5-x))?

The range of a function is the set of all possible output values. In this case, the range of the function f(x) = √(x(5-x)) is all non-negative real numbers, or [0, ∞). This is because the square root of a non-negative number is always non-negative.

Similar threads

Back
Top