Question

`f(x) = (x+1) (x+2)^(1/3), x>=-2` .

Answer 1

\[\text{Given:} \hspace{0.167em} f\left( x \right) = \left( x + 1 \right) \left( x + 2 \right)^\frac{1}{3} \]

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

\[\text { For the local maxima or minima, we must have }\]

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

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

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

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

\[ \Rightarrow x + 1 = - 3x - 6\]

\[ \Rightarrow x = \frac{- 7}{4}\]

Thus, `x = (- 7)/4` is the possible point of local maxima or local minima.

\[\text { Now,} \]

\[f''\left( \frac{- 7}{4} \right) = \frac{2}{3} \left( x + 2 \right)^\frac{- 2}{3} - \frac{2}{9}\left( x + 1 \right) \left( x + 2 \right)^\frac{- 5}{3} \]

\[\text { At } x = \frac{- 7}{4}: \]

\[ f''\left( \frac{- 7}{4} \right) = \frac{2}{3} \left( \frac{- 7}{4} + 2 \right)^\frac{- 2}{3} - \frac{2}{9}\left( \frac{- 7}{4} + 1 \right) \left( \frac{- 7}{4} + 2 \right)^\frac{- 5}{3} = \frac{2}{3} \left( \frac{1}{4} \right)^\frac{- 2}{3} + \frac{1}{18} \left( \frac{1}{4} \right)^\frac{- 5}{2} > 0\]

\[\text { So}, x = \frac{- 7}{4} \text { is the point of local minimum }. \]

\[\text { The local minimum value is given by}\]

\[f\left( \frac{- 7}{4} \right) = \left( \frac{- 7}{4} + 1 \right) \left( \frac{- 7}{4} + 2 \right)^\frac{1}{3} = \frac{- 3}{4} \left( \frac{1}{4} \right)^\frac{1}{3} = \frac{- 3}{4^\frac{4}{3}}\]