How do you solve and write the following in interval notation: #x^3 - 38x^2 <0#?

1 Answer
Aug 4, 2016

Answer:

Solution is #x<38#, except #x=0#

Explanation:

#x^3-38x^2<0# can be factorized as

#x^2(x-38)<0#

Now, if #x<0#, #x^2(x-38)<0# as #x^2# is positive, but #(x-38)# is negative and their product too is negative and inequality is satisfied.

If #0 < x < 38#, while #x^2# is positive, #(x-38)# is negative and their product too is negative and inequality is satisfied.

But if #x>38#, #x^2# and #(x-38)# are positive, and their product too is positive, and inequality is not satisfied.

Hence, solution is #x<38#, but #x=0# is not one as at #x=0#, #x^3-38x^2<0# is not true.