JEE Advance - Mathematics (2008 - Paper 1 Offline - No. 22)
The total number of local maxima and local minima of the function
$$f(x) = \left\{ {\matrix{ {{{(2 + x)}^3},} & { - 3 < x \le - 1} \cr {{x^{2/3}},} & { - 1 < x < 2} \cr } } \right.$$ is
$$f(x) = \left\{ {\matrix{ {{{(2 + x)}^3},} & { - 3 < x \le - 1} \cr {{x^{2/3}},} & { - 1 < x < 2} \cr } } \right.$$ is
0
1
2
3
Explanation
Given that, $$f(x) = \left\{ {\matrix{ {{{(2 + x)}^3},} & { - 3 < x \le - 1} \cr {{x^{2/3}},} & { - 1 < x < 2} \cr } } \right.$$
$$f'(x) = \left\{ {\matrix{ {3{{(2 + x)}^2},} & { - 3 < x < - 1} \cr {{2 \over 3}{x^{{{ - 1} \over 3}}},} & { - 1 < x < 2} \cr } } \right.$$
Clearly, $$f'(x)$$ changes signs from positive to negative as x passes through x $$=-1$$
$$f'(x) < 0$$ for all $$x\in(-3,-1)$$
Also, $$f'(x)>0$$ for all $$x\in$$
$$f'(x) < 0$$ for all $$x\in(-1,0)$$
But $$f'(0)$$ does not exist
So, $$f(x)$$ attains a local minimum at x = 0
Hence, the total number of local maxima and local minima is 2.
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