Solve the inequality in terms of intervals and illustrate the solution set on the real number line: (x^3)+3x<(4x^2)

x³ + 3x < 4x²

Subtract 4x² from both sides:

x³ - 4x²+ 3x < 0

Factor out an x:

x(x² - 4x + 3) < 0

Finish the factoring:

x(x - 1)(x - 3) < 0

Find the zeros of this inequality:  x = 0, x = 1, and x = 3:

This divides the number line into four open-ended regions: (-∞,0), (0,1), (1,3), (3,∞)

(The regions are open-ended because the problem has a '<' sign; if there were a '≤' sign, square brackets would be used with the numbers, but not the infinity signs.)

Now test each region. Chose one number in the region: if it works, that region is part of your answer; if it doesn't work, that region is not part of your answer. You cannot choose an endpoint of the region!

For (-∞,0):  I can choose any number in this region, I'll select -1 and test it:

--->  (-1)(-1 - 1)(-1 - 3)  =  (-1)(-2)(-3)  =  -6. This is less than 0, so this region works.

For (0,1):  I can choose any number in this region, I'll select .5 and test it:

--->  (.5)(.5 - 1)(.5 - 3)  =  (.5)(-.5)(-2.5)  =  .625. This is greater than 0, so this region doesn't work.

For (1,3):  I can choose any number in this region, I'll select 2 and test it:

--->  (2)(2 - 1)(2 - 3)  =  (2)(1)(-1)  =  -2. This is less than 0, so this region works.

For (3, ∞):  I can choose any number in this region, I'll select 4 and test it:

--->  (4)(4 - 1)(4 - 3)  =  (4)(3)(1)  =  12. This is greater than 0, so this region doesn't work.

Final answer:  (-∞,0) ∪ (1,3)

For graphing purposes, place open circles at 0, 1, and 3 and shade the region to the left of zero and between 1 and 3.