I need calculus help

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 Let f(x) = -x^4 - 3x^3 + 2x - 1 

Find the open intervals on which f is concave up

Find the open intervals on which f is concave down

Determine the x-coordinates of all inflection points of f

Apr 8th, 2015

The derivative of f(x) is f '(x) = - 4x^3 - 9x^2 + 2.

The second derivative of f(x) is f ''(x) = -12x^2 - 18x = -6x(2x + 3). Note that f ''(x) = 0 at the points x = 0 and 

x = -3/2. Since f ''(x) < 0 for x > 0 and for x < -3/2, the intervals on which f is concave down are (- infty, - 3/2) and   (0, + infty).

Since f ''(x) > 0 for -3/2 < x < 0, the interval on which f is concave up is (-3/2, 0).

The inflection points are at x = 0 and x = -3/2 where the f '' changes its sign.

Apr 8th, 2015

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