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

1 Answer
Jun 9, 2018

Answer:

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}#