What is the domain and range of f(x) =sqrt(( x- (3x^2)))?

1 Answer
Bill Jorgensen
Jun 9, 2018

Domain {x|x in RR:0<=x<=1/3}

Range {y|y in RR:0<=y<=sqrt3/6}

Explanation:

f(x) =sqrt(( x- (3x^2)))

Numbers under a radical must be greater than or equal to 0 or they are imaginary, so to solve the domain:

x- (3x^2)>=0

x- 3x^2>=0

x(1- 3x)>=0

x>=0

1-3x>=0

-3x>=-1

x<=1/3

So our domain is:

{x|x in RR:0<=x<=1/3}

Since the minimum input is sqrt0=0 the minimum in our range is 0.

To find the maximum we need to find the max of -3x^2+x

in the form ax^2+bx+c

aos = (-b)/(2a) = (-1)/(2*-3)=1/6

vertex (max) = (aos, f(aos))

vertex (max) = (1/6, f(1/6))

f(x)=-3x^2+x

f(1/6)=-3(1/6)^2+1/6=1/12

vertex (max) = (1/6, 1/12)

Finally, don't forget the square root, we have a maximum at x=1/6 of sqrt(1/12) =sqrt3/6 so our range is:

{y|y in RR:0<=y<=sqrt3/6}

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