Solve the Inequality x^3-12x^2>=37x+35?

Answer 1

x ≥ 14.68 (done with a graphing calculator)
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Ideas for doing it without a graphing calculator:
f(x) = x^3-12x^2-37x-35
f'(x) = 3x^2-24x-37
Solve f'(x) = 0 for x,
x = -1.323, 9.323
Since f(-1.323)<0 and f(9.323)<0, f(x) can have at most one root at x>9.323.

Answer 2

The proper method is to solve the corresponding equation first. x^3-12x^2=37x+35. Of course this forces one to solve a cubic equation. Several methods may be used for this.

Depending on what form you need the answer to be in determines which method you would use. If you need exact solutions for interval endpoints, an advanced algebra text may contain the formulae for solving cubic equations. If you do not need exact solutions for endpoints then much easier methods are available.

The easiest method by far would be to use a graphing calculator to plot y=x^3 - 12x^2 - 37x -35. Use the calculator's trace function to estimate the values of the x-intercepts, and report the intervals where the graph is on or above the x-axis.

Answer 3

You can solve this in a graphic way.

x^3 >= 12x^2 + 37x+35

Graphic 1) y = x^3 and 2) y = 12x^2 + 37 x +35

The first graphic is like that:

.......... ! .. o
.......... | .. o
_____o______
.........o |
....o..... |
...o.......|

The other one is a parabola. Find the points where the cubic function is above the quadratic one

Ilusion

Answer 4

The politically correct method is to remove the "greater than" symbol and solve for x. That being said, the expedient method is to solve it as it stands by trial-and-error. x is about 20.