Category Archives: Differentiation and its applications [O level]

Equations without stationary points

Given y = 4/√(2x-3), x > 1.5

(a)Find dy/dx

(b)Explain why the graph of y = 4/√(2x – 3) does not have a stationary point

(c)State whether y = 4√(2x – 3) is an increasing or decreasing function. Explain your answer clearly.

y = 4(2x – 3)-1/2

dy/dx = -2(2x – 3)-3/2(2)

           = -4(2x – 3)-3/2

(b) Since, dy/dx = -4(2x – 3)-3/2  , (2x-3) is always positive

     (2x – 3 )-3/2 > 0

  • dy/dx < 0, dy/dx ≠ 0, therefore graph has no stationary point

Maximum/Minimum gradient of a curve

Find the minimum gradient of the curve y = 2x3 – 9x2 + 5x + 3 and the value of x when the minimum gradient occurs.

y = 2x3 – 9x2 + 5x + 3

gradient, m = 6x2 – 18x + 5
dm/dx = 12x – 18

minimum gradient,

dm/dx = 0

12x – 18 =0

X = 3/2 occurs at minimum gradient

= 6(3/2)2 – 18(3/2) + 5

= -17/2

Maximum point and minimum point 1

A curve with equation in the form of y = ax + b/x 

y = ax + b/x2 has a stationary point at (3 , 4), where a and b are constants. Find the value of a and of b.

X = 3     y = 4

4 = 3a + b/a —–(1)

Y = ax + bx-2

dy/dx = a – 2bx-3—–(2)

sub dy/dx = 0, x = 3 into (2)

0 = a – 2b(3)-3

0 = a – 2b/27—–(3)

Sub (3) into (1)

4 = 3(2b/27) + b/9

4 = 1/3b

b = 12  ,   a = 2(12)/27

b = 12  ,   a = 8/9