Find the points at which the function $f$ given by $f(x)=(x-2)^{4}(x+1)^{3}$ has
$(i)$ local maxima
$(ii)$ local minima
$(iii)$ point of inflexion

(A) The given function is $f(x)=(x-2)^{4}(x+1)^{3}$.
First,we find the derivative $f'(x)$ using the product rule:
$f'(x) = 4(x-2)^{3}(x+1)^{3} + 3(x+1)^{2}(x-2)^{4}$
$f'(x) = (x-2)^{3}(x+1)^{2} [4(x+1) + 3(x-2)]$
$f'(x) = (x-2)^{3}(x+1)^{2} (4x + 4 + 3x - 6)$
$f'(x) = (x-2)^{3}(x+1)^{2} (7x - 2)$.
Setting $f'(x) = 0$,we get critical points $x = 2$,$x = -1$,and $x = \frac{2}{7}$.
Analyzing the sign of $f'(x)$ around these points:
$1$. For $x = 2$: $f'(x)$ changes from negative to positive as $x$ crosses $2$,so $x = 2$ is a point of local minima.
$2$. For $x = \frac{2}{7}$: $f'(x)$ changes from positive to negative as $x$ crosses $\frac{2}{7}$,so $x = \frac{2}{7}$ is a point of local maxima.
$3$. For $x = -1$: $f'(x)$ does not change sign as $x$ crosses $-1$ (since $(x+1)^{2}$ is always non-negative),so $x = -1$ is a point of inflexion.
Thus:
$(i)$ Local maxima at $x = \frac{2}{7}$.
$(ii)$ Local minima at $x = 2$.
$(iii)$ Point of inflexion at $x = -1$.