How do you find the intervals where the function $f(x)=(2x-1^2)(x-3)^2$ is concave up and concave down?

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Answer 1 · 2015-05-02T18:21:59

Concave down in $(-oo,13/6)$ and concave up in $(13/6,oo)$

For finding the intervals where f(x) is concave up or concave down, first get f'' (x).

f(x) is $(2x-1)(x-3)^2 = 2x^3 - 13x^2 +24x -9$

f'(x)= $6x^2-26x +24$ and f'' (x)= 12x-26. The inflection point is at x= $13/6$

For $x<=13/6$ , f''(x) is negative, hence f(x) would be concave down in the interval $(-oo, 13/6)$

For $x>= 13/6$ , f''(x) is positive, hence f(x) would be concave up in the interval $(13/6, oo)$

How do you find the intervals where the function f(x)=(2x-1^2)(x-3)^2f(x)=(2x-1^2)(x-3)^2 is concave up and concave down?