Question 12.7: Find any maximum points, minimum points and points of inflex......
Find any maximum points, minimum points and points of inflexion of y = x³ + 2x².
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 \text { or } x=-\frac{4}{3}. sing the second-derivative test we find 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^{\prime \prime}=0 \text { when } x=-\frac{2}{3}.
Since y” is negative when Since y” is negative when x<-\frac{2}{3} \text {, then } y^{\prime}, then y’ is decreasing there, that is the function is concave down. Also, y” is positive when x>-\frac{2}{3} \text { and so } y^{\prime} 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=x^3+2 x^2 is shown in Figure 12.13.
