Let f(x) = |(x - 1)(x^2 - 2x - 3)| + x - 3, x in R. If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to ____________.

f(x) = | (x - 1)(x + 1)(x - 3) | + (x - 3) f(x) = \ matrix (x - 3)(x^2) 3 le x le 4 cr (x - 3)(2 - x^2) 1 le x f'(x) = \ matrix 3x^2 - 6x 3 f'(3^ + ) > 0 f'(3^ - ) f'(1^ + ) > 0 f'(1^ - ) x in (1,3) f'(x) = 0 at one point -> Maximum x in (3,4) f'(x) ne 0 x in (0,1) f'(x) ne 0 So, 3 points.

Let fx x 1x2 2x 3 x 3x R If m and M are respectively th... | null (Mathematics)

f(x) = \left| {(x - 1)(x + 1)(x - 3)} \right| + (x - 3)

f(x) = \left\{ {\matrix{ {(x - 3)({x^2})} & {3 \le x \le 4} \cr {(x - 3)(2 - {x^2})} & {1 \le x < 3} \cr {(x - 3)({x^2})} & {0 < x < 1} \cr } } \right.

f'(x) = \left\{ {\matrix{ {3{x^2} - 6x} & {3 < x < 4} \cr { - 3{x^2} + 6x + 2} & {1 < x < 3} \cr {3{x^2} - 6x} & {0 < x < 1} \cr } } \right.

f'({3^ + }) > 0\,\,\,f'({3^ - }) < 0 \to

Minimum

f'({1^ + }) > 0\,\,\,f'({1^ - }) < 0 \to

Minimum

x \in (1,3)\,\,f'(x) = 0

at one point

\to

Maximum

x \in (3,4)\,\,f'(x) \ne 0

x \in (0,1)\,\,f'(x) \ne 0

So, 3 points.

Explanation

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