Stationary points of an implicit cubic

C

Total marks: 9

Solution

(a)

Implicit differentiation

Show that any stationary points on $C$ lie on $y=3x^2$ .

Explore this idea

Hint

Differentiate implicitly.
A stationary point has $\frac{dy}{dx}=0$ .

Answer · (a)

y=3x^2

Explore the idea

Flat tangents sit on y = 3x² (shown for A = 1)

Jump to the two stationary points; both satisfy y = 3x².

Position along the curve(0.61536, 1.13601)

Try an example

Tangent gradient

Stationary point on the right: $x\approx0.616$ and $y=3x^2\approx1.14$ . The tangent is horizontal and the curve peaks here.

The tangent is horizontal and the normal is vertical. Both axes use equal scales.

X³ + y³ − xy = 1
Tangent

Key idea

A flat tangent forces $y=3x^2$ ; the curve fixes where on that parabola the point lies.

Ideas to try

Check that each marked point satisfies

y=3x^2

(b) coordinates

Quadratics in a power

Find, in terms of $A$ , the $x$ -coordinates of the stationary points.

Hint

Substitute $y=3x^2$ into the curve.
The result is a quadratic in $x^3$ .

Answer · (b) coordinates

x_\pm=\frac13\sqrt[3]{1\pm\sqrt{1+27A}}

(b) range of A

Discriminant

Hence determine the range of values of $A$ for which $C$ has two distinct stationary points.

Explore this idea

Hint

Two distinct values of $x^3$ need a positive discriminant: $1+27A>0$ .

Answer · (b) range of A

A>-1/27

Explore the idea

How many x-values give 27x⁶ − 2x³ = A?

Each intersection supplies an algebraic stationary candidate. The question excludes A = 0.

Height of the horizontal linek = 0.1

Try an example

Test boundary heights

Buttons use full-precision boundary values; numbers shown here are rounded.

Number of distinct stationary candidates

Two stationary candidates for the nonzero parameter A.

All listed intersections are within the original drawing window. A touching point counts once; hollow endpoints are excluded.

Coordinates are rounded to 6 significant figures. Inspect a row to highlight that intersection and bring it into view.
Point	x	y
	-0.324611	0.1
	0.47663	0.1

27x⁶ − 2x³
Y = A
X-coordinates of stationary points
Intersection selected in the coordinate table

Key idea

The sextic's minimum value is $-\frac1{27}$ (at $x=\frac13$ ): below that there are none.

Ideas to try

Find the lowest

A

that still gives a stationary point.

(c)

Second derivative test

Determine the nature of each stationary point.

Explore this idea

Hint

Differentiate $(3y^2-x)\frac{dy}{dx}=y-3x^2$ again.
Put $\frac{dy}{dx}=0$ and $y=3x^2$ .
Use the sign of $\frac{d^2y}{dx^2}$ .

Answer · (c)

Plus root: maximum. Minus root: minimum.

Explore the idea

A maximum and a minimum on the same curve

Jump to each stationary point and look at which way the curve bends.

Position along the curve(-0.54169, 0.88027)

Try an example

Tangent gradient

Minimum: $\frac{d^2y}{dx^2}=\frac6{\sqrt{1+27A}}>0$ . The curve bends up and the tangent lies below it.

The tangent is horizontal and the normal is vertical. Both axes use equal scales.

X³ + y³ − xy = 1
Tangent

Key idea

Positive second derivative: minimum. Negative: maximum. The sign comes entirely from the cube root sign.

Ideas to try

On which side of the tangent does the curve lie at each point?

Stationary points of an implicit cubic

Solution

Implicit differentiation

Differentiate

Solve for $\frac{dy}{dx}$

Stationary: $\frac{dy}{dx}=0$

Flat tangents sit on y = 3x² (shown for A = 1)

Quadratics in a power

Substitute

Quadratic in $x^3$

Take cube roots

Discriminant

Discriminant

Exclude $A=0$

How many x-values give 27x⁶ − 2x³ = A?

Second derivative test

Differentiate again

At a stationary point

Plus root

Minus root

A maximum and a minimum on the same curve

Differentiate

Solve for dydx\frac{dy}{dx}dxdy​

Stationary: dydx=0\frac{dy}{dx}=0dxdy​=0

Substitute

Quadratic in x3x^3x3

Take cube roots

Discriminant

Exclude A=0A=0A=0

Differentiate again

At a stationary point

Plus root

Minus root

Solve for $\frac{dy}{dx}$

Stationary: $\frac{dy}{dx}=0$

Quadratic in $x^3$

Exclude $A=0$