Question
Let $$f\left( x \right)$$ be a function defined as below:
\[f\left( x \right) = \left\{ \begin{array}{l}
\sin \left( {{x^2} - 3x} \right),\,x \le 0\\
6x + 5{x^2},\,x > 0
\end{array} \right.\]
Then at $$x = 0,\,f\left( x \right)$$
A.
has a local maximum
B.
has a local minimum
C.
is discontinuous
D.
none of these
Answer :
has a local minimum
Solution :
$$\eqalign{
& {\text{In }} - \in < x \leqslant 0,\,f'\left( x \right) = \cos \left( {{x^2} - 3x} \right).\left( {2x - 3} \right) < 0 \cr
& \therefore \,f\left( x \right)\,{\text{is decreasing}}{\text{.}} \cr
& {\text{In }}x > 0,\,f'\left( x \right) = 6 + 10x > 0 \cr} $$
$$\therefore \,f\left( x \right)$$ is increasing. So, at $$x = 0,\,f\left( x \right)$$ has a local minimum.
$$\therefore $$ at $$x = 0,\,f\left( x \right)$$ is continuous because $$f\left( {0 - \in } \right) = f\left( 0 \right) = f\left( {0 + \in } \right)$$