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प्रश्न
Find the local minimum and local maximum of y = 2x3 – 3x2 – 36x + 10.
उत्तर
y = 2x3 – 3x2 – 36x + 10
`"dy"/"dx"` = 6x2 – 6x – 36 = 6(x2 – x – 6)
`"dy"/"dx"` = 0 gives 6(x2 – x – 6) = 0
6(x – 3) (x + 2) = 0
x = 3 (or) x = -2
`("d"^2"y")/"dx"^2` = 6(2x – 1)
Case (i): when x = 3,
`(("d"^2"y")/"dx"^2)_(x=3)`= 6(2 × 3 – 1)
= 6 × 5
= 30, positive
Since `("d"^2"y")/"dx"^2` is positive y is minimum when x = 3.
The local minimum value is obtained by substituting x = 3 in y.
Local minimum value = 2(33) – 3(32) – 36(3) + 10
= 2(27) – (27) – 108 + 10
= 27 – 98
= -71
Case (ii): when x = -2,
`(("d"^2"y")/"dx"^2)_(x=-2)`= 6(-2 × 2 – 1)
= 6 × -5
= -30, negative
Since `("d"^2"y")/"dx"^2` is negative, y is maximum when x = -2.
Local maximum value = 2(-2)3 – 3(-2)2 – 36(-2) + 10
= 2(-8) – 3(4) + 72 + 10
= -16 – 12 + 82
= -28 + 82
= 54
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