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nignag [31]
2 years ago
5

The square root of a negative number, such as StartRoot negative 144 EndRoot, is undefined. Explain why the square root of –x, S

tartRoot negative x EndRoot, is not necessarily undefined and what this means about the domain and range of f(x) = StartRoot negative x EndRoot.
Mathematics
1 answer:
monitta2 years ago
6 0

Part A

The reason why the (√(-x)) is not necessarily undefined, is because x is a placeholder, such that x can take on a negative value for which  (-x) is a positive number, therefore, √(-x) = √(+ve number) is a defined real number.

Part B

The domain is x ≤ 0

The range is f(x) ≥ 0

<h3>What is Real Number?</h3>

A real number is a value of a continuous quantity that can represent a distance along a line.

Here, The given and required information:

Given that √(-144) = Undefined, why is √(-x) not necessarily undefined

The given function, f(x) = √(-x)

Part A

The square root of a real number with a negative value is an imaginary number

The square root of a real number with a positive value or zero, is a real number

√(144) = ±12

A variable, x, is an input value of a function that is a quantity place holder in the function which is a function argument

The given function is f(x) = √(-x)

Where;

x = An argument of the function

When x ≤ 0, such that x is a negative real number or zero, we have;

x = R⁻ or 0

Therefore;

-x= -R⁻ or 0 = R⁺ or 0

-x = R⁺

From which we get;

f(-x)= √(-x) = √(R⁺) = Defined

f(-x) = √(R⁺) ≥ 0 The range of the function is f(x) ≥ 0

An example, when x = -16, we have;

-x = -(-16) = 16

√(-x) √(16) = 4 (Defined)

Part B;

Thee domain where the function is defined is where x = -R⁻,  x ≤ 0

The output of the function f(x) is defined which is the range = f(x) ≥ 0

Thus, the reason why the √(-x) is not  necessarily undefined, is because the value of x can represent a negative real number or zero.

Learn more about real numbers here;

brainly.com/question/17201233

#SPJ1

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