# How do you find the points of inflection if f'(x)= 12x^3 + 3x^2 + 42x?

Mar 29, 2015

The points of inflections are the points in which the second derivatives changes sign. The second derivative is the derivative of the first derivative, so we have
$f ' ' = \frac{d}{\mathrm{dx}} f ' = 36 {x}^{2} + 6 x + 42$
To find the zeroes of this function, we must impose
$36 {x}^{2} + 6 x + 42 = 0$.
Dividing both terms by 6, we can simplify the expression into
$6 {x}^{2} + x + 7 = 0$. This is the expression of a parabola, whose discriminant is negative (namely, ${b}^{2} - 4 a c = 1 - 4 \cdot 6 \cdot 7 = - 167$), and so it has no zeroes. This means that the second derivative is always positive, and thus the function has no points of inflections.

If you take a look at the graph, you'll see that the result is believable: if $f ' \left(x\right) = 12 {x}^{3} + 3 {x}^{2} + 42 x$, by integration you have that
$f \left(x\right) = 3 {x}^{4} + {x}^{3} + 24 {x}^{2} + c$, where $c$ is the integration constant (which only translates the function and thus doesn't change its shape). As you can see, the function never changes from being concave (concave downward) to convex (concave upward), or vice versa, which means that it has no points of inflection.
Here's the plot