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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.
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