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

Answer 1

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

#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.

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Answer 2

Kennedy Adams

To solve and write ( x^3 - 38x^2 <0 ) in interval notation, the solution is ( x \in (0, 38) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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