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

1.	Find the local minimum and local maximum of y = 2x3 – 3x2 – 36x + 10
Answer» y = 2x³ – 3x² – 36x + 10 \(\frac{dy}{dx}\) = 6x² – 6x – 36 = 6(x² – x – 6) \(\frac{dy}{dx}\) = 0 gives 6(x² – x – 6) = 0 6(x – 3) (x + 2) = 0 x = 3 (or) x = -2 \(\frac{d^2y}{dx^2}\) = 6(2x – 1) Case (i): when x = 3, \((\frac{d^2y}{dx^2})_{x=3}\) = 6(2 x 3 – 1) = 6 x 5 = 30, positive Since \(\frac{d^2y}{dx^2}\) 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, \((\frac{d^2y}{dx^2})_{x=-2}\) = 6(-2 x 2 – 1) = 6 x -5 = -30, negative Since \(\frac{d^2y}{dx^2}\) 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

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

Answer»

y = 2x³ – 3x² – 36x + 10

\(\frac{dy}{dx}\) = 6x² – 6x – 36 = 6(x² – x – 6)

\(\frac{dy}{dx}\) = 0 gives 6(x² – x – 6) = 0

6(x – 3) (x + 2) = 0

x = 3 (or) x = -2

\(\frac{d^2y}{dx^2}\) = 6(2x – 1)

Case (i): when x = 3,

\((\frac{d^2y}{dx^2})_{x=3}\) = 6(2 x 3 – 1)

= 6 x 5

= 30, positive

Since \(\frac{d^2y}{dx^2}\) 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,

\((\frac{d^2y}{dx^2})_{x=-2}\) = 6(-2 x 2 – 1)

= 6 x -5

= -30, negative

Since \(\frac{d^2y}{dx^2}\) 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