Question

Find the intervals in which the following function is strictly increasing or strictly decreasing. Also find the points of local maximum and local minimum, if any : f(x) = $(x-1{)}^3(x+2{)}^2$.

Answer 1

Given $f\left(x\right)=(x-1{)}^3(x+2{)}^2{f}^{\prime }\left(x\right)=(x-1{)}^2(x+2\left)\right(5x+4)$ For critical points, $f^{\prime }\left(x\right)=0\Rightarrow (x-1{)}^2(x+2\left)\right(5x+4)=0\therefore x=1,-2,-\frac{4}{5}$
$\begin{array}{ccc}\text{Interval}& \text{Sing of f'(x)}& f\left(x\right)is\\ (-\infty ,-2)& \text{Positive}& \text{Strictly increasing}\\ (-2,-\frac{4}{5})& \text{Negative}& \text{Strictly increasing}\\ (-\frac{4}{5},1)& \text{Positive}& \text{Strictly increasing}\\ (1,\infty )& \text{Positive}& \text{Strictly increasing}\end{array}$
Hence f(x) is strictly increasing in $(-\infty ,-2)$ and $(-\frac{4}{5},\infty )$ & strictly decreasing in $(-2,-\frac{4}{5})$.
Also in the left neighbourhood of -2, f'(x) is positive and in right neighbourhood of -2, f'(x) is negative and f'(-2) = 0. Therefore by the first derivative test, x= -2 is a point of local maximum. Also f'(x) changes its sign from negative to positive as x passes through $-\frac{4}{5}$ so, x = $-\frac{4}{5}$ is a point of local minimum.