Find the intervals where fx x4 2x3 36x2 150 is concave up

Find the intervals where f(x) = x4 + 2x3 -36x2 +150 is concave up and concave down. any points of inflection. Use the following information to sketch the graph of f(x) =2x2+1 / x2-4 find the vertical asymptotes

Solution

f(x) = x^4 + 2x^3 - 36x^2 + 150 f\'(x) = 4x^3 + 6x^2 - 72x f\'\'(x) = 12x^2 + 12x - 72 for inflection point, f\'\'(x) = 0 12x^2 + 12x - 72 = 0 x^2 + x - 6 = 0 x = -3,2 are the inflection point for concave up f\'\'(x) > 0 x^2 + x - 6 > 0 (x+3)(x-2)>0 x > 2, x > -3.....concave up -3 < x < 2........concave down