pre-calc help badly needed!

[chitchatkitkat14]

I need help with these 2 problems:

1. Solve, graph your answer on a number line, and then express your answer using interval notation.

22x+5x^2-3x^3 > 9x^3+16x^2-18x


2. Determine when a cube with a sida of x units will have the numerical value of its volume larger than the numerical value of its surface area.

 

[Erik_D]

solving inequalities

1) 22x+5x^2-3x^3 > 9x^3+16x^2-18x
First, I would subtract 22x+5x^2-3x^3 from both sides. Then factor this--it won't be too hard because one factor is x. After that, use a sign chart to determine when it will be negative.

1-solution) 22x+5x^2-3x^3 > 9x^3+16x^2-18x
0>12x^3+11x^2-40x
0>(x)(12x^2+11x-40)
so it would be 0 at [-11-sqrt(2064)]/24, 0, and [-11+sqrt(2064)]/24.
These are approximated to be -2.341, 0, and 1.424.
I used the numbers -3, -1, 1, and 2 to test signs on a sign chart.
The signs at these points are -, +, -, and + respectively.
It was negative (0>x) for points less than [-11-sqrt(2064)]/24 and points between 0 and [-11+sqrt(2064)]/24.
Interval notation would be (-infinity, [-11-sqrt(2064)]/24)U(0, [-11+sqrt(2064)]/24). To show it on a number line, darken these two areas, being sure to leave open circles on the ends to show that x cannot equal zero.

2) Because the volume of the cube is x^3 and the surface area is 6x^2, simply solve the inequality x^3>6x^2.

2-solution) x^3>6x^2
x^3-6x^2>0
x^2 (x-6) >0
It equals 0 at x=0, 0, and 6. Because the length must be a positive number, I will only use a sign chart evaluating at x=1 and 10.
I got - and + respectively, so the volume of the cube will be larger than the surface area when x>6.
Interval notation would be (6, infinity)