Differentiation

Edexcel IAL WMA12/01 /P2/Jan 2021/Q2 (Differentiation Stationary Points)

A curve has equation​​ 

y = x3 - x2 - 16x + 2 

• Using calculus, find the x coordinates of the stationary points of the curve.​​ 

(4)​​ 

• Justify, by further calculus, the nature of all of the stationary points of the curve.​​ 

(3)

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 = x3 - x2 - 16x + 2

dydx=3x2-2x-16

At stationary point,​​ dydx=0

3x2-2x-16=0

3x2-2x-16=0

Solving quadratic equation by breaking th emiddle term method.​​ 

3x-8x+2=0

3x2-x-16=0

3x2+6x-8x-16=0

3xx+2-8x+2=0

x+23x-8=0

3x-8=0                x+2=0

x=83                         x=-2

Hence, the x-cordinates of x are ​​ 83 & -2.

b-​​ 

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

dydx=3x2-2x-16

Diffrentiating the expression of​​ dydx,​​ ​​ to find the​​ d2ydx2

d2ydx2=6x-2

Substituting the value​​ x=-2 in d2ydx2​​ expression.

 d2ydx2=6-2=-14<0

Hence,​​ the stationary point at​​ x=-2​​ is local maximum.​​ 

Now, substituting​​ x=83 ,

d2ydx2=6283-2=16-2=14>0

Thus,​​ the stationary point at​​ x=83​​ is local minimum.​​