Concavity and Points of Inflection – A Level Maths Year 2

1 What the Second Derivative Tells You

The first derivative $\mathrm{f}'(x)$ is the gradient. The second derivative $\mathrm{f}''(x)$ is the rate of change of the gradient — it tells you which way the curve is bending.

If $\mathrm{f}''(x) > 0$ the gradient is increasing, so the curve bends upwards like a cup (∪). If $\mathrm{f}''(x) < 0$ the gradient is decreasing, so the curve bends downwards like a cap (∩).

Term	Condition	Shape	Chords
Convex	$\mathrm{f}''(x) \geqslant 0$ for every $x$ in the range	Bends upwards (∪)	Lie above the curve
Concave	$\mathrm{f}''(x) \leqslant 0$ for every $x$ in the range	Bends downwards (∩)	Lie below the curve

Memory aid: a concave curve is like the roof of a cave, ∩. Convex curves "hold water". The link to chords is why concavity decides whether the trapezium rule over- or underestimates.

2 Finding Where a Curve Is Convex or Concave

Differentiate twice to find $\mathrm{f}''(x)$.
Solve the inequality $\mathrm{f}''(x) > 0$ (convex) or $\mathrm{f}''(x) < 0$ (concave).
Give the answer as an inequality in $x$, e.g. "convex for $x > 2$".

Worked Example 1 — A cubic

$\mathrm{f}(x) = x^3 - 6x^2 + 9x + 1$. Find the set of values of $x$ for which the curve is concave, and the point where it changes from concave to convex.

①$\mathrm{f}'(x) = 3x^2 - 12x + 9$, $\;\mathrm{f}''(x) = 6x - 12$.

②Concave: $6x - 12 < 0 \Rightarrow x < 2$. Convex for $x > 2$.

③The change happens at $x = 2$, where $\mathrm{f}(2) = 8 - 24 + 18 + 1 = 3$: the point $(2, 3)$ — see the graph above.

For any cubic the point of inflection lies exactly half-way between the two stationary points: here $x = 2$ is half-way between $x = 1$ and $x = 3$.

Worked Example 2 — Showing a function is convex everywhere

Show that $\mathrm{f}(x) = \mathrm{e}^{2x} + x^2$ is convex for all real $x$.

①$\mathrm{f}'(x) = 2\mathrm{e}^{2x} + 2x$, $\mathrm{f}''(x) = 4\mathrm{e}^{2x} + 2$.

②$\mathrm{e}^{2x} > 0$ for all $x$, so $\mathrm{f}''(x) > 2 > 0$ for all $x$. Hence $\mathrm{f}$ is convex everywhere.

Worked Example 3 — A condition on a constant

Find the values of $k$ for which $y = x^4 + kx^3 + 6x^2$ is convex for all $x$.

①$y'' = 12x^2 + 6kx + 12$.

②We need $12x^2 + 6kx + 12 \geqslant 0$ for all $x$: a positive quadratic that never goes below the axis, so its discriminant must satisfy $b^2 - 4ac \leqslant 0$.

③$36k^2 - 4(12)(12) \leqslant 0 \Rightarrow 36k^2 \leqslant 576 \Rightarrow k^2 \leqslant 16$.

④$-4 \leqslant k \leqslant 4$.

3 Points of Inflection

Point of inflection

A point where the curve changes from concave to convex, or from convex to concave:
$\mathrm{f}''(x) = 0$ and $\mathrm{f}''(x)$ changes sign either side

$\mathrm{f}''(x) = 0$ is not enough. For $y = x^4$, $y'' = 12x^2$ is zero at $x = 0$ but positive on both sides. The curve is convex throughout, and $(0, 0)$ is a minimum, not a point of inflection. Always check the sign change.

A point of inflection may or may not be a stationary point. If $\mathrm{f}'(x) = 0$ there as well, it is a stationary point of inflection (like $y = x^3$ at the origin); otherwise it is a non-stationary point of inflection, where the tangent crosses the curve.

Worked Example 4 — Two kinds of inflection

Find the points of inflection on $y = x^4 - 4x^3$ and say whether each is stationary.

①$y' = 4x^3 - 12x^2 = 4x^2(x - 3)$, $y'' = 12x^2 - 24x = 12x(x - 2)$.

②$y'' = 0$ at $x = 0$ and $x = 2$.

③Sign of $y''$: positive for $x < 0$, negative for $0 < x < 2$, positive for $x > 2$. It changes sign at both, so both are points of inflection.

④At $x = 0$: $y = 0$ and $y' = 0$ — a stationary point of inflection at $(0, 0)$.

⑤At $x = 2$: $y = 16 - 32 = -16$ and $y' = 16(-1) = -16 \neq 0$ — a non-stationary point of inflection at $(2, -16)$.

Worked Example 5 — An exponential product

Find the point of inflection on $y = x\mathrm{e}^x$ and state where the curve is convex.

①$y' = \mathrm{e}^x + x\mathrm{e}^x = (1 + x)\mathrm{e}^x$.

②$y'' = \mathrm{e}^x + (1 + x)\mathrm{e}^x = (2 + x)\mathrm{e}^x$.

③$\mathrm{e}^x > 0$, so the sign of $y''$ is the sign of $2 + x$: it changes at $x = -2$. Point of inflection $\left(-2, -2\mathrm{e}^{-2}\right)$.

④Convex for $x > -2$; concave for $x < -2$.

Worked Example 6 — Two inflections

Show that $y = \ln(1 + x^2)$ has points of inflection at $x = \pm1$, and find the set of values of $x$ for which it is convex.

①$y' = \dfrac{2x}{1 + x^2}$ (chain rule).

②Quotient rule: $y'' = \dfrac{(1 + x^2)(2) - 2x(2x)}{(1 + x^2)^2} = \dfrac{2(1 - x^2)}{(1 + x^2)^2}$.

③The denominator is positive, so $y''$ has the sign of $1 - x^2$: positive for $-1 < x < 1$, negative for $|x| > 1$. It changes sign at $x = \pm1$, giving points of inflection $(\pm1, \ln 2)$.

④Convex for $-1 < x < 1$ (strictly; or $-1 \leqslant x \leqslant 1$ using $\mathrm{f}'' \geqslant 0$).

4 When the Second Derivative Test Fails

From Year 1: at a stationary point, $\mathrm{f}'' > 0$ means a minimum and $\mathrm{f}'' < 0$ a maximum. If $\mathrm{f}'' = 0$ the test gives no information — it could be a maximum, minimum or point of inflection. Check the sign of $\mathrm{f}'$ on either side instead.

Worked Example 7 — A stationary point of inflection

Find and classify the stationary point of $y = x^3 + 3x^2 + 3x$.

①$y' = 3x^2 + 6x + 3 = 3(x + 1)^2 = 0 \Rightarrow x = -1$, $y = -1 + 3 - 3 = -1$.

②$y'' = 6x + 6 = 0$ at $x = -1$: inconclusive.

③$y' = 3(x + 1)^2 \geqslant 0$ on both sides: the curve rises, levels off, then rises again.

④$y''$ changes sign at $x = -1$, so $(-1, -1)$ is a stationary point of inflection. (Indeed $y = (x + 1)^3 - 1$.)

5 Quick Reference

Convex

$\mathrm{f}''(x) \geqslant 0$; bends upwards; chords above.

Concave

$\mathrm{f}''(x) \leqslant 0$; bends downwards; chords below.

Inflection

$\mathrm{f}''(x) = 0$ and $\mathrm{f}''$ changes sign.

Stationary inflection

Also $\mathrm{f}'(x) = 0$, e.g. $y = x^3$ at $0$.

Not an inflection

$y = x^4$: $\mathrm{f}''(0) = 0$ but no sign change.

Max / min test

$\mathrm{f}'' < 0$ max, $\mathrm{f}'' > 0$ min, $\mathrm{f}'' = 0$ check signs.

Factor out

Positive factors like $\mathrm{e}^x$ don't affect the sign of $\mathrm{f}''$.

Cubics

Inflection half-way between the turning points.

6 Practice Questions

Question 1

Find the coordinates of the point of inflection on $y = x^3 - 3x^2 + 4$.

▶ Show solution

$y' = 3x^2 - 6x$, $y'' = 6x - 6 = 0 \Rightarrow x = 1$; $y''$ changes sign there.

$y = 1 - 3 + 4 = 2$: point of inflection $(1, 2)$.

Question 2

Find the set of values of $x$ for which $\mathrm{f}(x) = x^4 - 6x^2$ is concave.

▶ Show solution

$\mathrm{f}''(x) = 12x^2 - 12 = 12(x - 1)(x + 1)$.

Concave when $\mathrm{f}''(x) \leqslant 0$: $-1 \leqslant x \leqslant 1$, i.e. $\{x : -1 \leqslant x \leqslant 1\}$.

Question 3

Show that $y = x\ln x$ is convex for all $x > 0$.

▶ Show solution

$y' = \ln x + x \cdot \tfrac1x = \ln x + 1$, $\;y'' = \tfrac1x$.

For $x > 0$, $\tfrac1x > 0$, so the curve is convex.

Question 4

For both $y = x^5$ and $y = x^6$, the second derivative is zero at $x = 0$. Determine in each case whether the origin is a point of inflection.

▶ Show solution

$y = x^5$: $y'' = 20x^3$, negative for $x < 0$ and positive for $x > 0$. Sign change: a (stationary) point of inflection.

$y = x^6$: $y'' = 30x^4 \geqslant 0$ on both sides. No sign change: not a point of inflection (it is a minimum).

Question 5

Find the point of inflection on $y = 2\sin x$ for $0 < x < 2\pi$, and state where the curve is concave.

▶ Show solution

$y'' = -2\sin x = 0 \Rightarrow x = \pi$ (in the range). $y'' < 0$ for $0 < x < \pi$ and $y'' > 0$ for $\pi < x < 2\pi$, so there is a sign change.

Point of inflection $(\pi, 0)$. Concave for $0 < x < \pi$.

Question 6

Find the exact $x$-coordinates of the points of inflection on $y = x^2\mathrm{e}^{-x}$.

▶ Show solution

$y' = 2x\mathrm{e}^{-x} - x^2\mathrm{e}^{-x} = (2x - x^2)\mathrm{e}^{-x}$.

$y'' = (2 - 2x)\mathrm{e}^{-x} - (2x - x^2)\mathrm{e}^{-x} = (x^2 - 4x + 2)\mathrm{e}^{-x}$.

$x^2 - 4x + 2 = 0 \Rightarrow x = 2 \pm \sqrt2$. These are simple roots, so $y''$ changes sign at each: both are points of inflection.

Question 7

Find and classify the stationary points of $y = x^3 + 3x^2 - 2$, and find the range of values of $x$ for which the curve is convex.

▶ Show solution

$y' = 3x^2 + 6x = 3x(x + 2) = 0 \Rightarrow x = 0$ or $x = -2$.

$y'' = 6x + 6$. At $x = 0$: $y'' = 6 > 0$, minimum $(0, -2)$. At $x = -2$: $y'' = -6 < 0$, maximum $(-2, 2)$.

Convex when $6x + 6 > 0$, i.e. $x > -1$.

Question 8

Find and classify the stationary point on $y = x^4 - 4x^3$ at $x = 3$, and explain why the second derivative test cannot classify the one at $x = 0$.

▶ Show solution

$y' = 4x^2(x - 3)$, $y'' = 12x(x - 2)$.

At $x = 3$: $y = 81 - 108 = -27$ and $y'' = 36 > 0$: minimum $(3, -27)$.

At $x = 0$: $y'' = 0$, so the test is inconclusive. $y' = 4x^2(x - 3)$ is negative on both sides of $0$, so it is a stationary point of inflection (as in Worked Example 4).

Question 9

The curve $y = x^3 + ax^2 + bx$ has a point of inflection at $(1, -4)$. Find $a$ and $b$.

▶ Show solution

$y'' = 6x + 2a$. At $x = 1$: $6 + 2a = 0 \Rightarrow a = -3$.

The point lies on the curve: $1 - 3 + b = -4 \Rightarrow b = -2$.

(A cubic's $y''$ is linear and always changes sign at its root, so this is indeed a point of inflection.)

Question 10

The shape of a normal distribution curve is given by $\mathrm{f}(x) = \mathrm{e}^{-x^2/2}$.

(a) Show that $\mathrm{f}''(x) = (x^2 - 1)\mathrm{e}^{-x^2/2}$.
(b) Find the exact coordinates of the points of inflection.
(c) State the set of values of $x$ for which the curve is convex.
(d) Find the equation of the tangent at the point of inflection with $x > 0$, and where it meets the $x$-axis.

▶ Show solution

(a) $\mathrm{f}'(x) = -x\mathrm{e}^{-x^2/2}$ (chain rule). Product rule: $\mathrm{f}''(x) = -\mathrm{e}^{-x^2/2} + (-x)(-x)\mathrm{e}^{-x^2/2} = (x^2 - 1)\mathrm{e}^{-x^2/2}$.

(b) $\mathrm{e}^{-x^2/2} > 0$, so $\mathrm{f}'' = 0$ when $x = \pm1$, with a sign change at each. Points of inflection $\left(\pm1, \mathrm{e}^{-1/2}\right)$.

(c) Convex when $x^2 - 1 > 0$: $\{x : x < -1\} \cup \{x : x > 1\}$.

(d) $\mathrm{f}'(1) = -\mathrm{e}^{-1/2}$. Tangent: $y - \mathrm{e}^{-1/2} = -\mathrm{e}^{-1/2}(x - 1)$, i.e. $y = \mathrm{e}^{-1/2}(2 - x)$. It meets the $x$-axis at $x = 2$.

In statistics this is why the points of inflection of a normal curve lie exactly one standard deviation either side of the mean.

Concavity and Points of Inflection · A Level Maths Year 2 · Created with MathJax