Differentiation

Edexcel IAL WMA12/01 /P2/Jan 2020/Q10 (Differentiation Stationary Points)

A curve C has equation​​ 

y=4x3-9x+kx           x>0 

where k is a constant.​​ 

The point P with x coordinate​​ 12​​ lies on C.​​ 

Given that P is a stationary point of C,​​ 

• show that​​ k = -32​​ 

(4)​​ 

• Determine the nature of the stationary point at P, justifying your answer.​​ 

(2)

The curve C has a second stationary point.​​ 

• Using algebra, find the x coordinate of this second stationary point.​​ 

(4)

SOLUTION

a-​​ 

(At stationary points, the gradient of the curve is equal to zero. Therefore, we will diffrentiate the equation of curve and equate the diffrentiated expression to zero. )

y=4x3-9x1+kx-1

dydx=12x2-9-kx-2

dydx=12x2-9-kx2

At stationary point,​​ dydx=0. Thus, substituting the value​​ x=12.

 dydx= 0 

12 122-9-k122=0

12 14 -9-k24=0

3-9-4k=0

- 6=4k

k=-64

k=-32

b-​​ Differentiating the first derivative of​​ y​​ to find double derivative and then putting the value of x as​​ 12​​ in the double derivated expression to find the nature of​​ y​​ at​​ x=12.

(If a function​​ f(x)​​ has a stationary point when​​ x=a, then if​​ f’’a<0, the point is a local maximum)

We already have an expression for ​​ dydx.

dydx=12x2-9+32x-2 

On diffrentiating it second time.​​ 

d2ydx2=24x-3x-3

d2ydx2=24x-3x3

On substituting​​ x=12,

d2ydx2=24 12-3123

d2ydx2=12-318

d2ydx2=-12<0

Hence, the nature of the stationary point is maximum.

c-​​ 

To find the second stationary point, let us again equate the​​ ​​ expression to zero, but this time we also have k value to subtitute.

dydx=12x2-9+32x-2 

dydx=0

12x2-9+32x2=0

24x4-18x2+3=0

Let​​ b=x2, so the equation becomes​​ 

24b2-18b+3=0

8b2-6b+1=0

2b-14b-1=0

2b=1    4b=-1

b=12        b=14

Now, substituting back the value of​​ b=x2.

x2=12          x2=14

x= ±12      x= ±14

It is given in the question that​​ x>0, so we will choose the value of​​ x​​ greater than 0. Ignoring the negative values.​​ 

x=12                x=12

The x-coordinate of the second stationary point will be​​ x=12.

Don’t leave the answer with an irrational denominator, do rationalise the denominator.​​ 

x=12×22

x=22

 ​​​​ The x-coordinate of the second stationary point is​​ x=22.  ​​ ​​​​