Review question

# What can we say if $y = m(x-a)$ is tangent to $y = x^3 - x$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8287

## Solution

The graphs of $y = x^3 - x$ and $y = m(x-a)$ are drawn on the axes below. Here $m>0$ and $a \le -1$.

The line $y=m(x-a)$ meets the $x$-axis at $A = (a,0)$, touches the cubic $y=x^3 -x$ at $B$ and intersects again with the cubic at $C$. The $x$-coordinates of $B$ and $C$ are respectively $b$ and $c$.

1. Use the fact that the line and cubic touch when $x=b$, to show that $m=3b^2-1$.

For the line and the cubic to touch at $x=b$, the gradients must be equal at this point.

The gradient of the cubic is given by $\frac{dy}{dx} = 3x^2 - 1,$ and the gradient of the line is $m$.

Therefore, equating these expressions at $x=b$ gives $m = 3b^2 - 1.$

1. Show further that $a = \frac{2b^3}{3b^2 - 1}.$

The value of $y$ must be the same for each graph at $x=b$. Hence we have $b^3 - b = m(b-a).$

Now from part (i) we know that $m=3b^2 - 1$, so substituting this into the above expression gives $b^3 - b = (3b^2 - 1)(b-a).$

We can rearrange this to get $a$ by dividing by $3b^2-1$ to give $b-a=\frac{b^3 - b}{3b^2 - 1}$ and then finding \begin{align*} a &{}= b - \frac{b^3 - b}{3b^2 - 1}\\ &{}= \frac{b(3b^2-1) -(b^3 - b)}{3b^2 -1}\\ &{}= \frac{2b^3}{3b^2 - 1}. \end{align*}
1. If $a=-10^6$, what is the approximate value of $b$?

If $a$ is large and negative, the line $y=m(x-a)$ (which is still a tangent to the curve somewhere between $-1$ and $0$) will be almost horizontal.

So the point $B$ will lie almost at the maximum point of the curve, which is where $3x^2-1=0$, so $b\approx-\dfrac{1}{\sqrt{3}}$.

Another way of thinking about this is that the fraction in (ii) must be very large and negative.

For this, either $b$ must be very large and negative, or the denominator must be very close to zero.

But $b$ lies between $-1$ and $0$, so it must be the latter, and $3b^2-1\approx0$, giving $b \approx - \dfrac{1}{\sqrt{3}}$ (taking the negative square root).

1. Using the fact that $x^3 - x - m(x-a) = (x-b)^2 (x-c)$ (which you need not prove), show that $c=-2b$.

We can put both sides in a polynomial form, so that we can see the coefficients of $x^3$, $x^2$ etc. on each side. Doing this gives $x^3-(m+1)x +ma = x^3 -(2b + c)x^2 +(b^2 + 2bc)x -b^2 c.$

Since the statement is true for all $x$, it must be true that the coefficient of any power of $x$ must be the same on both sides.

In particular the coefficient of $x^2$ on the left side, $0$, must be equal to the coefficient on the right side, $-(2b + c)$. Therefore $2b+c=0$, and hence $c=-2b$.

1. $R$ is the finite region bounded above by the line $y=m(x-a)$ and bounded below by the cubic $y = x^3 - x$. For what value of $a$ is the area of $R$ largest?

Show that the largest possible area of $R$ is $\dfrac{27}{4}$.

If $a$ increases (getting closer to $-1$), then $b$ decreases and the line $y=m(x-a)$ rises, making the area of $R$ larger.

Therefore we want to choose the largest possible value of $a$ in order to make the area the largest. So as $a \le -1$, the area of $R$ is largest when $a=-1$.

And when $a=-1$, the line $y=m(x-a)$ must be tangent to the curve at the point $(-1,0)$ since this is a point on both the line and the curve for this value of $a$.

So $B$ coincides with $A$ and $b=-1$.

We can also deduce this algebraically using the earlier results, as follows. By part (ii), we see that $-1 = \frac{2b^3}{3b^2 - 1}.$

Rearranging this gives $2b^3 + 3b^2 - 1 = 0$, which can be factorised as $(b+1)^2 (2b-1) = 0$. Since $b$ can’t be positive, we conclude that $b=-1$.

Then using part (iv), we obtain $c=2$, and from part (i), we see that $m=3b^2 - 1 = 2$.

So we can now calculate \begin{align*} \text{Area of R} &{}= \int_b^c (m(x+1) - (x^3 - x)) \, dx \\ &{}= \int_{-1}^2 -x^3 + 3x + 2 \, dx \\ &{}= \left[ -\frac{x^4}{4} + \frac{3x^2}{2} + 2x \right]_{-1}^2 \\ &{}= 6-\left(-\frac{3}{4}\right)\\ &{}= \frac{27}{4}. \end{align*}