#### To determine

**To find: **The two points on the curve y=x4-2x2-x that have a common tangent line

#### Answer

The two points on the curve y=x4-2x2-x that have a common tangent line are,

(1, -2) and (c, d)

#### Explanation

**1) Concept:**

By calculating slope of the line at two points having common tangent and equating it to find values of both points

**2) Formula:**

i. Difference rule:

ddxfx-gx=ddxfx-ddx(gx)

ii. Power rule:

ddxxn=nxn-1

iii. Constant multiple rule:

ddxCfx=Cddx(fx)

iv. Point-slope formula:

y-y1=m(x-x1)

v. Standard formula:

a3-b3=(a-b)(a2-ab+b2)

**3) Calculations:**

Equation of curve is, y=x4-2x2-x

Consider two points (a, y(a) ) and (b, y(b) ) having a common tangent line

Consider equation of curve,

y=x4-2x2-x

Differentiate with respect to x,

dydx=ddx(x4-2x2-x)

By using difference rule,

dydx=ddx(x4)-ddx(2x2)-ddx(x)

By using constant multiple rule and power rule,

dydx=4x3-2ddx(x2)-1

By using power rule,

dydx=4x3-4x-1

Now dydx is the slope of the tangent line

The equation of tangent line at x = a is,

By using point-slope formula,

y-a4-2a2-a=(4a3-4a-1)(x-a)

⇒y=4a3-4a-1x-a-a4-2a2-a

⇒y=4a3-4a-1x+(-3a4+2a2)

Now at x = b,

⇒y=4b3-4b-1x+(-3b4+2b2)

Now at x=a and x=b we have same tangent line,

Therefore,

4a3-4a-1=4b3-4b-1 and -3a4+2a2=-3b4+2b2

The first equation gives,

a3-b3=a-b

By using formula,

⇒(a-b)(a2-ab+b2)=a3-b3

And assuming a≠b we have

⇒a2-ab+b2=1

The second equation gives,

3a4-b4=2(a2-b2)

By using formula,

⇒3a2-b2a2+b2=2(a2-b2)

Which is true if a = - b or 3a2+b2 =2

Suppose a = - b thenSubstituting into a2-ab+b2=1

Gives

1=a2-a2+a2

⇒a=±1

So that a = 1 and b = -1 and vice versa

Thus the points (1, -2) and (-1, 0) have common tangent line

**Conclusion:**

The two points on the curve y=x4-2x2-x that have a common tangent line are,

(1, -2) and (-1, 0)