What is the domain and range of this square root function?

Question

f(x)= -Sqr root((x^2) -9x) . If you didn't catch it says negative square root

Answer 1

DOMAIN:

This limiting factor in this domain is that the expression under the square root must be positive.

So, let's first find when it changes between positive and negative by setting that equal to zero:



x^2 - 9x = 0

x(x - 9) = 0

So, at x=0 and x=9, this expression equals zero.



Graphing x^2 - 9x = 0 or plugging in some test values for x lets us see that this parabola is only positive for x<0 and for x>9. Since you can only take the square root of a positive number or zero, your domain (or valid x-values) is ( -infinity, 0]U[9, infinity).



RANGE:

Since the square root of any number is always positive, and you're multiplying your square root by (-1), your range will span all negative y-values, including zero. We can write this range as ( -infinity, 0].

Answer 2

DOMAIN → Function → Range

f(x) = -sqrt[x^2-9x]



The domain is all values that x can take on. BUT you cannot have a negative inside the square root. SO,

Set the insides greater-than-or-equal-to zero, and solve. The result will be domain:

x^2-9x > 0

x-9> 0 or

x > 9

Then the domain is "all x > 9". OR Range is all x

Answer 3

I think that the answer is this

the number inside the root must be POSITIVE at all time and what gives us that result is or Zero

the Domain = (-infinity,0] U [9, +Infinity)

considering that the number will be always positive inside the root or a zero we will

then the range is all the out puts which will be

y <= 0

I suggest you check the answer

Answer 4

f(x)  =   -√[x² - 9x]

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Domain

Square roots are real only if the radicand is not negative.

So, the domain of f satisfies

             x² - 9x ≥ 0

              x(x-9) ≥ 0

                    ⇓

           x ≤ 0  OR  x ≥ 9        ← Domain



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Range

Since square roots are never negative, the least a square root can be is 0.

Other than that, the square root is always positive and increasing so that its

opposite,  -√ ̅ ̅   , must be either 0 or always negative and decreasing.



It follows that since x²-9x does equal zero for x=0 and x=9, we have that

f(x)  =   -√[x² - 9x]   continuously decreases from 0.

Therefore,

                 The Range of f(x) is  y ≤ 0



Have a good one!

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Answer 5

x^2-9x >= 0

x(x-9) >= 0

Domain x>=9 or x<=0

Range -infinity to 0