Question

f(x) = x³  (x \[-\] 1)² .

Answer 1

\[\text { Given }: f\left( x \right) = x^3 \left( x - 1 \right)^2 \]

\[ \Rightarrow f'\left( x \right) = 3 x^2 \left( x - 1 \right)^2 + 2 x^3 \left( x - 1 \right)\]

\[\text { For a local maximum or a local minimum, we must have }\]

\[ f'\left( x \right) = 0\]

\[ \Rightarrow 3 x^2 \left( x - 1 \right)^2 + 2 x^3 \left( x - 1 \right) = 0\]

\[ \Rightarrow x^2 \left( x - 1 \right)\left\{ 3x - 3 + 2x \right\} = 0\]

\[ \Rightarrow x^2 \left( x - 1 \right)\left( 5x - 3 \right) = 0\]

\[ \Rightarrow x = 0, 1, \frac{3}{5}\]

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 \[\left( \frac{3}{5} \right)^3 \left( \frac{3}{5} - 1 \right)^2 = \frac{27}{125} \times \frac{4}{25} = \frac{108}{3125}\]

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