Nonlinear Inequalities from Pre-Cal, anyone?

Question

Would you tell me a) how to solve these kinds of problems and b) how to factor them?
Directions: Solve the inequality and write the solution set in interval notation.

(2x^3) - (x^4) < or = to 0

Thank you!

Would you tell me a) how to solve these kinds of problems and b) how to factor them?
Directions: Solve the inequality and write the solution set in interval notation.

(2x^3) - (x^4) < or = to 0

Thank you!

Thanks guys. That helped. So the "interval notation" thing-that is referring to the line graph with the different sections, right?

Answer 1

How to factor:

Find the common factor. In this case x^3

2x^3 - x^4 = 0
(x^3)(2-x) = 0
x=0 or x=2

Now, you have two critical points, x=0 and x=2. You have 3 intervals and two points.

Evaluate EACH of the following for the condition of < or = in the orignal equation.

x < 0
x = 0
0 < x < 2
x = 2
x > 2

Answer 2

There are 3 aspects. x + 3 is damaging to the left of -3 and helpful to the stunning x - a million is damaging to the left of a million and helpful to the stunning a million/(x - 5) is damaging to the left of 5 and helpful to the stunning. we would like a damaging product, so we seem for periods with the two 3 negatives or one damaging and a pair of positives. If there are 0 or 2 damaging aspects, the negatives will cancel out and the effect would be greater effective than 0. start up on the farthest left. If x < -3, all 3 aspects are damaging, so the product is decrease than 0. it fairly is a factor of the answer set. between -3 and a million, basically x + 3 is helpful. the different 2 are damaging, so the product is larger than 0. it fairly is not area of the answer set. From a million to 5, basically a million/(x-5) is damaging. With one damaging and a pair of features, the product would be much less that 0. it is interior the answer. Above 5, all aspects are helpful, so it fairly is not area of the answer. e.) (-inf, -3) union (a million,5)

Answer 3

find it's roots, and find the value for x parts below the curve:
(x^3)((2x)-1)=0
x^3=0, x=0
2x-1=0
2x=1
x=.5
now plot y=(2x^3)-(x^4).
The part below the x-axis is the answer.