# Find intervals of concavity. f(x)= (x^3) - (x^2) + x – 3 (-3 ≤ x ≤ 3) Find intervals of concavity. f(x)= (x^3) - (x^2) + x – 3 (-3 ≤ x ≤ 3)

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You need to determine the first derivative and then you can find the second derivative and to solve the equation f''(x) = 0.

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

f''(x) = ( 3x^2 - 2x + 1)' => f''(x) = 6x - 2

You need to solve the equation f''(x) = 0 such that:

f''(x) = 0 => 6x - 2 = 0 => 6x = 2 => x = 2/6 => x = 1/3

Hence, the function has an inflection point at x = 1/3, thus the graph changes its concavity at = 1/3.

Notice that the second derivative is negative for x in [-3;1/3] and it is positive for x in [1/3 ; 3], hence, the graph of the function is concave down over (-3;1/3) and it is concave up over (1/3 ; 3), as the graph below proves it: