a couple of Qs (1 Viewer)

[pc4pc]

1. find any turning points on the curve y=(4x^2-1)^4 and determine their nature


2. the curve y = ax^3+bx^2-x+5 has a point of inflexion at (1, -2). find the values of a and b.

thankyouu:wave:

 

[nathan71088]

Q1,
y = (4x^2 - 1)^4
y = ((2x -1)^4)((2x+1)^4)
By using a multiplication rule as well as the chain rule:

dy = ((2x+1)^4)(4(2x - 1)^3)(2)) + ((2x -1)^4)(4(2x + 1)^3)(2))dx

dy = ((2x+1)^4)(8)(2x - 1)^3) + ((2x -1)^4)(8)(2x + 1)^3)
dx
=(8)((2x-1)^3)((2x+1)^3)(2x-1+2x+1)

=(8)((2x-1)^3)((2x+1)^3)(4x)

=(32x)((2x-1)^3)((2x+1)^3)

=(32x){[(2x-1)(2x+1)]^3}

then to check for turning points make the derivative = 0

(32x){[(2x-1)(2x+1)]^3} = 0

to expand this:

(32x)(2x-1)(2x+1)(2x-1)(2x+1)(2x-1)(2x+1) = 0

that means that one of these brackets = 0.

therefore:

32x = 0
or
(2x-1) = 0
or
(2x+1) = 0

therefore turning points at:

x = 0, -1/2, 1/2

for a second derivative to find the nature, differentiate again to get:

= (768x){[(2x-1)(2x+1)]^3}

then sub in the turning points to see whether they give a negative, positive answer or a "zero" answer. Tjis will tell you their nature.

 

[nathan71088]

Q2.

y = ax^3 + bx^2 - x +5 and (1,-2) is a point of inflexion. This means the second derivative = 0 when x = 1, and (1,-2) lies on the curve.

So,
y' = 3ax^2 + 2bx - 1
and
y'' = 6ax + 2b

therefore

0 = 6a(1) + 2b
0 = 6a +2b

and

-2 = a(1)^3 + b(1)^2 - (1) +5

-2 = a +b +4

-6 = a + b

a = -6 -b

subbing into the other equation with only a's and b's

0 = 6(-6-b) +2b

0 = -36 -6b +2b


4b = -36
b = -9
a = 3

 

[pc4pc]

cheeers =D you got em right !