What is the range of the function #f(x)= -sqrt((x^2) -9x) #?

2 Answers
Jul 8, 2017

Answer:

Range of #f(x) = (-oo, 0]#

Explanation:

#f(x) = -sqrt(x^2-9x)#

First let's consider the domain of #f(x)#

#f(x)# is defined where #x^2-9x>=0#

Hence where #x<= 0# and #x>=9#

#:.# Domain of #f(x) = (-oo, 0] uu [9, +oo)#

Now consider:

#lim_ (x->+-oo) f(x) =-oo#

Also: #f(0)= 0 # and #f(9) = 0#

Hence the range of #f(x) = (-oo, 0]#

This can be seen by the graph of #f(x) below.

graph{-sqrt(x^2-9x) [-21.1, 24.54, -16.05, 6.74]}

Jul 8, 2017

Answer:

Range : #f(x) <= 0#, in interval notation : #(-oo,0]#

Explanation:

#f(x) = - sqrt(x^2-9x)#

Range: Under root should be #>=0# , So #f(x) <=0#

Range : #f(x) <= 0#, in interval notation : #(-oo,0]#

graph{-(x^2-9x)^0.5 [-320, 320, -160, 160]} [Ans]