Question

f(x) = x³ (x$-$1)²

Answer 1

$$\begin{gathered} \mathrm{Given}:f\left(x\right)=x^3{\left(x-1\right)}^2 \\ \Rightarrow f\text{'}\left(x\right)=3x^2{\left(x-1\right)}^2+2x^3\left(x-1\right) \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maximum}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minimum},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 3x^2{\left(x-1\right)}^2+2x^3\left(x-1\right)=0 \\ \Rightarrow x^2\left(x-1\right)\left\{3x-3+2x\right\}=0 \\ \Rightarrow x^2\left(x-1\right)\left(5x-3\right)=0 \\ \Rightarrow x=0,1,\frac{3}{5} \end{gathered}$$



Since f '(x) changes from negative to positive when x increases through 1, x = 1 is the point of local minima.
The local minimum value of f (x) at x = 1 is given by
${\left(1\right)}^3{\left(1-1\right)}^2=0$

Since f '(x) changes from positive to negative when x increases through $\frac{3}{5}$, x = $\frac{3}{5}$ is the point of local maxima.
The local minimum value of f (x) at x = $\frac{3}{5}$ is given by
${\left(\frac{3}{5}\right)}^3{\left(\frac{3}{5}-1\right)}^2=\frac{27}{125}\times \frac{4}{25}=\frac{108}{3125}$

Since f '(x) does not change from positive as x increases through 0, x = 0 is a point of inflexion.

Disclaimer: The solution in the book is incorrect. The solution here is created according to the question given in the book.