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

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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.**