Economists spend a lot of time testing claims:

But how do we actually test whether a rule is true?

P.S roll up your sleeves and explore the question… once you’ve grappled with it for a while, if you are still stuck…Graduated hints below — but challenge yourself first! (Answer comes in the next edition! Subscribe to receive it directly to your inbox!)

£18.3 bn

This week’s marginal number is 1.3%.

Borrowing - the difference between tax receipts and government spending - was £18.3bn in August, almost a fifth higher than the year before, the Office for National Statistics (ONS) said.

That figure was £3.5bn more than official forecasters expected.

Debt interest from decisions in the past carry an opportunity cost of decisions in the present and future.

This is the awkward arithmetic of public finance: when debt interest rises, the government does not need to announce a new policy for the budget constraint to tighten. The constraint tightens by itself.

It is a reminder that the marginal pound matters. Every extra pound spent servicing debt is a pound that cannot be used elsewhere, unless taxes rise, borrowing rises further, or spending is cut.

Fiscal policy is not only about what governments choose to do next. It is also about the commitments they already carry.

Mo, here’s the final graph sketch for:

y = x³ / (x² − 1)

The solid blue curve is the function

x = −1, x = 1, and the oblique asymptote y = x.

Make sure it has these features:

If you want a more in-depth step-by-step answer, see below:

The trick is to split the curve into regions around the dangerous values of x, especially where the denominator is zero.

The function is undefined when the denominator is zero.

Denominator:

x² − 1 = 0

So:

x² = 1

Therefore:

x = ±1

So the domain is:

all real x except x = −1 and x = 1

Or:

x ∈ R, x ≠ −1, x ≠ 1

Vertical asymptotes occur where the denominator is zero, provided the numerator is not also zero there.

At x = −1:

Numerator:

x³ = (−1)³ = −1

Denominator:

x² − 1 = 0

So there is a vertical asymptote at:

x = −1

At x = 1:

Numerator:

x³ = 1³ = 1

Denominator:

x² − 1 = 0

So there is another vertical asymptote at:

x = 1

Therefore the vertical asymptotes are:

x = −1 and x = 1

The important thing is to look at the sign of the numerator and denominator.

The function is:

y = x³ / (x² − 1)

This means:

x → −1⁻

For values just less than −1, such as −1.1:

So:

negative / positive = negative

The denominator is getting very close to zero, so the value becomes very large in size.

Therefore:

as x → −1⁻, y → −∞

This means:

x → −1⁺

For values just greater than −1, such as −0.9:

So:

negative / negative = positive

The denominator is getting very close to zero, so the value becomes very large.

Therefore:

as x → −1⁺, y → +∞

This means:

x → 1⁻

For values just less than 1, such as 0.9:

So:

positive / negative = negative

The denominator is getting very close to zero.

Therefore:

as x → 1⁻, y → −∞

This means:

x → 1⁺

For values just greater than 1, such as 1.1:

So:

positive / positive = positive

The denominator is getting very close to zero.

Therefore:

as x → 1⁺, y → +∞

Now look at:

y = x³ / (x² − 1)

When x is very large, the dominant terms are:

x³ in the numerator

and:

x² in the denominator

So roughly:

y ≈ x³ / x²

which simplifies to:

y ≈ x

So as x → ∞, the curve behaves roughly like:

y = x

Therefore:

as x → ∞, y → ∞

And as:

x → −∞

the curve behaves roughly like:

y = x

So:

as x → −∞, y → −∞

But we can be more precise by doing algebraic division.

Divide:

x³ by x² − 1

We can rewrite the function as:

y = x + x / (x² − 1)

This is because:

x(x² − 1) = x³ − x

so:

x³ = x(x² − 1) + x

Therefore:

x³ / (x² − 1) = x + x / (x² − 1)

Now as x becomes very large, the leftover fraction:

x / (x² − 1)

gets closer and closer to zero.

So the oblique asymptote is:

y = x

There is no horizontal asymptote.

The curve approaches the line:

y = x

as x → ∞ and as x → −∞.

One nice extra detail:

Set x = 0.

y = 0³ / (0² − 1)

y = 0 / −1

y = 0

So the y-intercept is:

(0, 0)

Set y = 0.

A fraction equals zero when its numerator is zero, provided the denominator is not zero.

So:

x³ = 0

Therefore:

x = 0

So the x-intercept is also:

(0, 0)

The curve passes through the origin.

The critical values are:

x = −1, x = 0, x = 1

These split the real line into four intervals.

Example: x = −2

So:

y is negative

Therefore the curve is below the x-axis when:

x < −1

Example: x = −0.5

So:

negative / negative = positive

Therefore the curve is above the x-axis when:

−1 < x < 0

Example: x = 0.5

This is the “squaring numbers between 0 and 1 makes them smaller” point.

For example:

0.5² = 0.25

So:

x² − 1 < 0

Therefore:

positive / negative = negative

So the curve is below the x-axis when:

0 < x < 1

Example: x = 2

So:

y is positive

Therefore the curve is above the x-axis when:

x > 1

The function is:

y = x³ / (x² − 1)

Using the quotient rule:

dy/dx = [(x² − 1)(3x²) − x³(2x)] / (x² − 1)²

Now expand the numerator:

dy/dx = [3x²(x² − 1) − 2x⁴] / (x² − 1)²

dy/dx = [3x⁴ − 3x² − 2x⁴] / (x² − 1)²

dy/dx = (x⁴ − 3x²) / (x² − 1)²

Factorise the numerator:

dy/dx = x²(x² − 3) / (x² − 1)²

Stationary points occur where:

dy/dx = 0

So:

x²(x² − 3) = 0

This gives:

x² = 0

so:

x = 0

or:

x² − 3 = 0

so:

x = ±√3

So the stationary points occur at:

x = −√3, x = 0, x = √3

Substitute into:

y = x³ / (x² − 1)

Since:

x² = 3

and:

x³ = −3√3

we get:

y = −3√3 / (3 − 1)

y = −3√3 / 2

So one stationary point is:

(−√3, −3√3 / 2)

This is approximately:

(−1.73, −2.60)

We already know:

y = 0

So:

(0, 0)

Since:

x² = 3

and:

x³ = 3√3

we get:

y = 3√3 / (3 − 1)

y = 3√3 / 2

So another stationary point is:

(√3, 3√3 / 2)

This is approximately:

(1.73, 2.60)

The derivative is:

dy/dx = x²(x² − 3) / (x² − 1)²

The denominator is always positive where the function exists, because it is squared.

So the sign of the derivative depends on:

x²(x² − 3)

Since x² is always positive or zero, the main factor is:

x² − 3

So:

Therefore:

So:

local maximum: (−√3, −3√3 / 2)

stationary point of inflection: (0, 0)

local minimum: (√3, 3√3 / 2)

It feels odd that the local maximum has a negative y-value, but that is perfectly possible. It is the highest point on that particular branch between x = −∞ and x = −1.

The curve has three branches.

There is a useful shortcut too.

The function is:

f(x) = x³ / (x² − 1)

Now test f(−x):

f(−x) = (−x)³ / ((−x)² − 1)

f(−x) = −x³ / (x² − 1)

So:

f(−x) = −f(x)

That means the function is odd.

So the graph has rotational symmetry about the origin.

This is why the right-hand branch and left-hand branch mirror each other through the origin.

For:

y = x³ / (x² − 1)

Domain:

x ≠ −1, x ≠ 1

Vertical asymptotes:

x = −1 and x = 1

Oblique asymptote:

y = x

Intercept:

(0, 0)

Stationary points:

local maximum at (−√3, −3√3 / 2)

stationary point of inflection at (0, 0)

local minimum at (√3, 3√3 / 2)

Limits near asymptotes:

as x → −1⁻, y → −∞

as x → −1⁺, y → +∞

as x → 1⁻, y → −∞

as x → 1⁺, y → +∞

As x becomes very large:

as x → ∞, y → ∞, approaching y = x

as x → −∞, y → −∞, approaching y = x

The curve is a lovely test because it combines asymptotes, sign analysis, odd and even powers, fractions, and oblique asymptotes all in one deceptively simple function

Marginal gains… Answer for today’s question is out in two days… Come back then or subscribe to get it direct to your inbox!

P.S.

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