Question

How do you find points of inflection and determine the intervals of concavity given $y = 4{x^3} + 21{x^2} + 36x - 20$?

Answer 1

To solve this we first need to find the first derivative then the second derivative. After finding the second derivative we need to equate it to zero and we find the value of ‘x’. After that to find the concavity we split the interval $( - \infty ,\infty )$ into $( - \infty ,x) \cup (x,\infty )$. We find the value of the second derivative at each two intervals. If the second derivative is positive then it is concave up or if it is negative then it is concave down.

Complete step by step answer:
Given $y = 4{x^3} + 21{x^2} + 36x - 20$.
Differentiate with respect to ‘x’ we have,
$y' = 12{x^2} + 42x + 36$
Differentiate again with respect to ‘x’ we have,
$y'' = 24x + 42$.
Now equate double integration to zero,
$24x + 42 = 0$
Simplifying we have,
$24x = - 42$
Dividing 24 on both sides we have,
$x = - \dfrac{{42}}{{24}}$
$\Rightarrow x = - 1.75$.
Thus there is an inflection at $x = - 1.75$.
Now to find the concavity we split the interval $( - \infty ,\infty )$ into $( - \infty , - 1.75) \cup ( - 1.75,\infty )$
Now take the interval $( - \infty , - 1.75)$.
We take a value in this interval.
Since $- 2 \in ( - \infty , - 1.75)$
Put $x = - 2$in the $y'' = 24x + 42$
$y'' = - 48 + 42$
$\Rightarrow y'' = - 6$
Thus we have negative values in the interval $( - \infty , - 1.75)$.
Thus for every $x \in ( - \infty , - 1.75)$ the value of the second derivative is negative. Hence ‘y’ is concave down in the interval$( - \infty , - 1.75)$ .
Now take the interval $( - 1.75,\infty )$.
Since $1 \in ( - 1.75,\infty )$
Put $x = 1$in the$y'' = 24x + 42$.
$y'' = 24(1) + 42$
$\Rightarrow y'' = 24 + 42$
$\therefore y'' = 66$
Thus we have positive value in the interval $( - 1.75,\infty )$. Thus for every $x \in ( - 1.75,\infty )$ the value of the second derivative is positive. Hence ‘y’ is concave up in the interval $( - 1.75,\infty )$ .

Thus we have inflection at $x = - 1.75$ and concave up in the interval $( - 1.75,\infty )$ and concave down in the interval $( - \infty , - 1.75)$.

Note: Follow the same procedure for these kinds of problems. We have open intervals not the closed interval. In the open interval $( - 1.75,\infty )$ we do not include the value -1.75. If we have a closed interval instead of open we can take the value -1.75. By taking the second derivative we can tell if slope is increasing or decreasing. If the second derivative is positive then slope is increasing. If the second derivative is negative then slope is decreasing. In other words concave upward is when slope is increasing and concave downward is when slope is decreasing.