Question
Let $$f\left( x \right) = {x^3} + 3{x^2} + 3x + 2.$$ Then, at $$x = - 1$$
A.
$$f\left( x \right)$$ has a maximum
B.
$$f\left( x \right)$$ has a minimum
C.
$$f'\left( x \right)$$ has a maximum
D.
$$f'\left( x \right)$$ has a minimum
Answer :
$$f'\left( x \right)$$ has a minimum
Solution :
$$\eqalign{
& f\left( x \right) = {\left( {x + 1} \right)^3} + 1 \cr
& \therefore \,f'\left( x \right) = 3{\left( {x + 1} \right)^2} \cr
& f'\left( x \right) = 0\,\, \Rightarrow x = - 1 \cr
& {\text{Now, }}f'\left( { - 1 - \in } \right) = 3{\left( { - \in } \right)^2} > 0,\,f'{\left( { - 1 + \in } \right)^2} = 3{ \in ^2} > 0 \cr} $$
$$\therefore \,f\left( x \right)$$ has neither a maximum nor a minimum at $$x = - 1.$$
$$\eqalign{
& {\text{Let }}f'\left( x \right) = \phi \left( x \right) = 3{\left( {x + 1} \right)^2} \cr
& \therefore \,\phi '\left( x \right) = 6\left( {x + 1} \right) \cr
& \phi '\left( x \right) = 0\,\, \Rightarrow x = - 1 \cr
& \phi '\left( { - 1 - \in } \right) = 6\left( { - \in } \right) < 0,\,\,\phi '\left( { - 1 + \in } \right) = 6 \in > 0 \cr} $$
$$\therefore \,\phi \left( x \right)$$ has a minimum at $$x = - 1.$$