Question

determine the intervals on which the function is concave upwards or concave downwards

Answer 1

Given the function f(x) defined as:

$f(x)=5x^3-30x^2-7x-2$

The change in concavity is given by the inflection points of the function. These points are calculated by the solutions of the equation f''(x) = 0. Then, taking the second derivative of f(x):

$f^{\prime}^{\prime}(x)=30x-60$

Now, the inflection points are:

$\begin{gathered} f^{\prime}^{\prime}(x)=0 \\ 30x-60=0 \\ x=2 \end{gathered}$

There is only one inflection point at x = 2. Then, analyzing the concavity for x < 2 and x > 2:

$\begin{gathered} f^{\prime\prime}(0)=30\cdot0-60=-60\text{ (Downwards)} \\ f^{\prime\prime}(4)=30\cdot4-60=60\text{ (Upwards)} \end{gathered}$

Then, the intervals are:

$\begin{gathered} (-\infty,2)\to Downwards \\ (2,+\infty)\to Upwards \end{gathered}$