Question

Find the local minimum and local maximum of y = 2x³ – 3x² – 36x + 10.

Answer 1

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