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प्रश्न
Find the maximum and minimum of the following functions : y = 5x3 + 2x2 – 3x.
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उत्तर १
y = 5x3 + 2x2 – 3x
∴ `dy/dx = d/dx(5x^3 + 2x^2 - 3x)`
= 5 × 3x2 + 2 × 2x – 3 × 1
= 15x2 + 4x – 3
and `(d^2y)/(dx^2) = d/dx(15x^2 + 4x - 3)`
= 15 × 2x + 4 x 1 – 0
= 30x + 4
`dy/dx` = 0 gives 15x2 + 4x – 3 = 0
∴ 15x2 + 9x – 5x – 3 = 0
∴ 3x(5x + 3) – 1(5x + 3) = 0
∴ (5x + 3)(3x – 1) = 0
∴ x = `-(3)/(5) or x = (1)/(3)`
∴ the roots of `dy/dx = 0` are `x_1 = -(3)/(5) and x_2 = (1)/(3)`.
Second Derivative Test:
(a) `((d^2y)/dx^2)_("at" x = - 3/5) = 30(-3/5) + 4` = – 14 < 0
∴ by the second derivative test, y is maximum at x = `-(3)/(5)` and maximum value of y at x = `-(3)/(5)`
= `5(-3/5)^3 + 2(-3/5)^2 - 3(-3/5)`
= `(-27)/(25) + (18)/(25) + (9)/(5)`
= `(36)/(25)`
(b) `((d^2y)/dx^2)_("at" x = 1/3) = 30(1/3) + 4` = 14 > 0
∴ by the second derivative test, y is minimum at x = `(1)/(3)` and minimum value of y at x = `(1)/(3)`
= `5(1/3)^3 + 2(1/3)^2 - 3(1/3)`
= `(5)/(27) + (2)/(9) - 1`
= `-(16)/(27)`
Hence, the function has maximum value `(36)/(25)` at x = `-(3)/(5)` and minimum value `-(16)/(27) "at" x = (1)/(3)`.
उत्तर २
y = 5x3 + 2x2 – 3x
∴ `dy/dx = d/dx(5x^3 + 2x^2 - 3x)`
= 5 × 3x2 + 2 × 2x – 3 × 1
= 15x2 + 4x – 3
`dy/dx` = 0 gives 15x2 + 4x – 3 = 0
∴ 15x2 + 9x – 5x – 3 = 0
∴ 3x(5x + 3) – 1(5x + 3) = 0
∴ (5x + 3)(3x – 1) = 0
∴ x = `-(3)/(5) or x = (1)/(3)`
∴ the roots of `dy/dx = 0` are `x_1 = -(3)/(5) and x_2 = (1)/(3)`.
First Derivative Test:
(a) `dy/dx` = 15x2 + 4x – 3 = (5x + 3)(3x – 1)
Consider x = `-(3)/(5)`
Let h be a small positive number. Then
`(dy/dx)_("at" x = -3/5 - h) = [5(-3/5 - h) + 3][3 (-3/5 - h) - 1]`
= `( - 3 - 5h + 3)(-9/5 - 3h - 1)`
= `-5h (-14/5 - 3h)`
= `5h (14/5 + 3h) > 0`
and `(dy/dx)_("at" x = -3/5 + h) = [5(-3/5 + h) + 3][3 (-3/5 + h) - 1]`
= `(- 3 + 5h + 3)(-9/5 + 3h - 1)`
= `5h(3h - 14/5) < 0`,
as h is small positive number.
∴ by the first derivative test, y is maximum at x = `-(3)/(5)` and maximum value of y at x = `-(3)/(5)`
= `5(-3/5)^3 + 2(-3/5)^2 - 3(-3/5)`
= `-(27)/(25) + (18)/(25) + (9)/(5) = (36)/(25)`
(b) `dy/dx` = 15x2 + 4x – 3 = (5x + 3)(3x – 1)
Consider x = `(1)/(3)`
Let h be a small positive number. Then
`(dy/dx)_("at" x =1/3 - h) = [5(1/3 - h) + 3][3 (1/3 - h) - 1]`
= `(5/3 - 5h + 3)(1 - 3h - 1)`
= `(14/5 - 5h)(-3h)` < 0, as h is small positive number
and `(dy/dx)_("at" x = 1/3 + h) = [5(1/3 + h) + 3][3 (1/3 + h) - 1]`
= `(5/3 + 5h + 3)(1 + 3h - 1)`
= `(14/3 + 5h)(3h) > 0`
∴ by the first derivative test, y is minimum at x = `(1)/(3)` and minimum value of y at x = `(1)/(3)`
= `5(1/3)^3 + 2(1/3)^2 - 3(1/3)`
= `(5)/(27) + (2)/(9) - 1`
= `(-16)/(27)`
Hence, the function has maximum value `(36)/(25) "at" x = -(3)/(5)` and minimum value `-(16)/(27) "at" x = (1)/(3)`.
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