Verify if x^3-3x+2>0 for x>1.

We'll have to prove that the graph of the function x^3 -
3x + 2 is above x axis if x > 1.

We'll calculate the
1st derivative of the function:

f'(x) = 3x^2 -
3

We'll cancel the
derivative:

3x^2 - 3 = 0

We'll
divide by 3:

x^2 - 1 = 0

We'll
recognize the difference of two squares:

(x-1)(x+1) =
0

We'll cancel each factor;

x
- 1 = 0 => x = 1

x + 1 = 0 => x =
-1

Since the expression of the first derivative is positive
over the range (1 , +infinite), then the function is strictly increasin over the range
(1 , +infinite) and f(1) represents a local minimum point of the
function.

f(1) = 1 - 3 + 2 =
0

Any value of the function is larger then the minimum
point:

f(x) = x^3 - 3x + 2 > f(1) =
0

The inequality x^3 - 3x + 2 > 0 is
verified if x>1.