What is the domain of the function: #f(x) =sqrt(x^2(x-3)(x-4))#?

2 Answers
Oct 2, 2015

Answer:

#D_(f(x)) = (-oo,3]uu[4,+oo)#

Explanation:

Given
#color(white)("XXX")f(x)=sqrt(x^2(x-3)(x-4))#

To find the domain we need to determine which values of #x# are not valid.

Since the #sqrt("negative value")# is undefined (for Real numbers)

#x^2(x-3)(x-4) >= 0#

#x^2 >= 0# for all #x in RR#
#(x-3) > 0# for all #x>3, in RR#
#(x-4) > 0# for all #x>4, in RR#

The only combination for which
#color(white)("XXX")x^2(x-3)(x-4) < 0#
is when #(x-3) > 0# and #(x-4) < 0#

That is the only non-valid values for (Real) #x# occur when
#color(white)("XXX")x > 3# and #x < 4#

Oct 2, 2015

Answer:

#(-oo, 3] uu [4, oo)#

Explanation:

The domain is where the radicand (the expression under the square root sign) is non-negative.

We know that #x^2 >= 0# for all #x in RR#.

So in order that #x^2(x-3)(x-4) >= 0#, we must either have #x^2 = 0# or #(x-3)(x-4) >= 0#.

When #x<=3#, both #(x-3) <= 0# and #(x-4)<=0#, so #(x-3)(x-4) >= 0#

When #3 < x < 4#, #(x-3) > 0# and #(x-4) < 0#, so #(x-3)(x-4) < 0#.

When #x >= 4#, both #(x-3)>=0# and #(x-4)>=0#, so #(x-3)(x-4) >= 0#.

So #x^2(x-3)(x-4)>=0# when #x in (-oo,3] uu [4, oo)#

Note that this domain already includes the point #x = 0#, so the #x^2 = 0# condition gives us no extra points for the domain.