## Q. 12.5

Find any maximum points, minimum points and points of inflexion of y = x³ + 2x².

## Verified Solution

Given y = x³ + 2x² then y′ = 3x² + 4x and y′′ = 6x + 4. Let us first find any maximum and minimum points. The first derivative y′ is zero when 3x² + 4x = x(3x + 4) = 0, that is when x = 0 or  $x=-\frac{4}{3}$.  Using the second-derivative test we nd y′′(0) = 4 which corresponds to a minimum point. Similarly,  $y^{\prime \prime}\left(-\frac{4}{3}\right)=-4$  which corresponds to a maximum point.

We seek points of inflexion by looking for points where y′′ = 0 and then examining the concavity on either side. y′′ = 0 when  $x=-\frac{2}{3}$.

Since y′′ is negative when  $x<-\frac{2}{3},$  then y′ is decreasing there, that is the function is concave down. Also, y′′ is positive when  $x>-\frac{2}{3}$  and so y′ is then increasing, that is the function is concave up. Hence there is a point of inflexion when  $x=-\frac{2}{3}$.  The graph of y = x³ + 2x² is shown in Figure 12.13.