how to solve 1

Question

1. Given: y= -2x^3 + 3x^2 +12x - 5 Determine using differentiation, the co-ordinates of the maximum and minimum turning points. 2. Given: y=10^x tan x Differentiate by the use of the product rule 3. Differentiate the following with respect to x: y= [(1-cos x)/sin x) - (3/cosec x) + (3/x) + (sqaure-root x) --------- 1. Differentiate from first principles if: F(x) = 3 (square root x^6) 2. Given: y=(2x^3 - 7) to the power 4 Differentiate by using chain rule. 3. Differentiate with respect to x: y=(5x^3) - (7/square root 3x) - 2p + (2/x^-3) - ln4x - cot 2 x 4. Given: y=x^3 + 3x^2 - 9x + 5 Calculate, with that aid of differentiation, the co-ordinates of the turning points.

Answer 1

1)

y= -2x^3 + 3x^2 +12x - 5

Take derivative both sides with respected to x.
f ' (x) = - 2(3x²) + 3(2x) + 12
        = - 6x² + 6x + 12.
Again derivative to each side with respect to x.
f " (x) = -6(2x) + 6.
         = -12x + 6.
To find the turning points., to make the first derivative equal to zero or f ' (x) = 0.
- 6x² + 6x + 12 = 0.
x² - x - 2 = 0.
x² - 2x + x - 2 = 0.
x(x - 2) + 1(x - 2) = 0.
(x - 2)(x + 1) = 0.
x + 1= 0 and x - 2 = 0.
x = - 1 and x = 2.

f " (c) > 0 (positive) ------> minimum point.
f " (c) < 0 (negative) ------> maximum point.

f " (x) = -12x + 6.
At x = -1

f " (-1) = -12(-1) + 6 = 18>0

At x = -1, f(x) is minimum.

Find the the minimum value to substitute x = -1 in f(x).

f(-1) = -2(-1)³ + 3(-1)² +12(-1) - 5

          = 2 +3 -12 -5

          = -12

At x = 2

f " (2) = -12(2) + 6 = -18<0

At x = 2, f(x) is maximum.

Find the the maximum value to substitute x = 2 in f(x).

f(-1) = -2(2)³ + 3(2)² +12(2) - 5

          = -16 +12 +24 -5

          = 15