# Given the function f(x)= 1/12x^4 + 1/6x^3-3x^2-2x+1 how do you find any points of inflection and determine where the curve is concave up or down?

Apr 21, 2015

If $f$ is twice differentiable on an open set A, the inflections are the points where ${f}^{' '}$ changes sing. This is a polynomial so it's obvious that it's smooth (i.e. infinitively differentiable) on $\mathbb{R}$, so here, the inflection points are the points where ${f}^{' '} = 0$ and exist $n > 1$ such that ${f}^{\left(2 n + 1\right)} \ne 0$.

So ${f}^{' '} = {x}^{2} + x - 6$ and its roots are
$\frac{- 1 \pm \sqrt{1 + 24}}{2} = \left\{2 , - 3\right\}$

Now we notice that the third derivative: $2 x + 1$ has zero only in 1, so it's not zero in the candidate inflection points, and this tells us that -3 is a falling point and 2 is a rising point

(this does make sense since we know how the graphic of a quartic polynomial behave)
Now, if the function is twice differentiable, the concavity is determined by the sign of its second derivative:
${f}^{' '} > 0 \in U : = \left\{x < - 3 \mathmr{and} x > 2\right\}$
${f}^{' '} < 0 \in D : = \left\{- 3 < x < 2\right\}$

So in U the curve is concave up, and in D the curve is concave down