| Type | Description |
|---|---|
| Local maximum | A peak — the curve rises to it, then falls |
| Local minimum | A trough — the curve falls to it, then rises |
| Point of inflection | Flat, but the curve continues in the same direction |
- Differentiate to get $\dfrac{\mathrm{d}y}{\mathrm{d}x}$.
- Set it equal to zero and solve for $x$.
- Substitute each $x$ back into the original equation to get $y$.
- State the answers as coordinates.
Find the stationary point of $y = x^2 - 8x + 3$.
Find the stationary points of $y = 2x^3 - 3x^2 - 12x + 5$.
$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0$ → maximum (curve bends downwards, like a cap)
$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0$ → inconclusive — use the gradient-sign test instead
Determine the nature of the stationary points of $y = 2x^3 - 3x^2 - 12x + 5$, found above at $(-1, 12)$ and $(2, -15)$.
$+ \to 0 \to -$ is a maximum · $- \to 0 \to +$ is a minimum
same sign both sides is a point of inflection
Find and classify the stationary point of $y = x^3$.
Classify the stationary point of $y = x^4$ at the origin.
| $x$ | $-1$ | $0$ | $1$ |
|---|---|---|---|
| $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x^3$ | $-4$ | $0$ | $4$ |
| Sign | negative | zero | positive |
- Find the stationary points and classify them.
- Find the $y$-intercept by setting $x = 0$.
- Find the $x$-intercepts by factorising, if possible.
- Decide the end behaviour from the leading term.
- Draw a smooth curve through everything, and label each feature.
Sketch $y = x^3 - 3x^2$, showing all key features.
Find them
Solve $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$.
Get $y$
Substitute into the original equation.
Give coordinates
Both $x$ and $y$.
Minimum
$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0$.
Maximum
$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0$.
Zero second derivative
Inconclusive — use the sign test.
Sign test
$+\,0\,-$ maximum; $-\,0\,+$ minimum.
Say it in words
State the conclusion, not just the number.
How many?
Quadratic: $1$. Cubic: $0$ or $2$.
Touching the axis
A repeated root is a stationary point.
Find the stationary point of $y = x^2 - 6x + 2$ and state its nature.
▶ Show solution
$2x - 6 = 0$, so $x = 3$ and $y = 9 - 18 + 2 = -7$.
$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 > 0$, so $(3, -7)$ is a minimum.
Find the stationary point of $y = 5 + 4x - x^2$ and state its nature.
▶ Show solution
$4 - 2x = 0$, so $x = 2$ and $y = 5 + 8 - 4 = 9$.
$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -2 < 0$, so $(2, 9)$ is a maximum.
Find the stationary points of $y = x^3 - 12x$.
▶ Show solution
$3x^2 - 12 = 0$, so $x^2 = 4$ and $x = \pm 2$.
At $x=2$: $y = 8 - 24 = -16$. At $x=-2$: $y = -8 + 24 = 16$.
Stationary points $(2, -16)$ and $(-2, 16)$.
Classify the stationary points found in Question 3.
▶ Show solution
$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6x$
At $x=2$: $12 > 0$, so $(2,-16)$ is a minimum.
At $x=-2$: $-12 < 0$, so $(-2,16)$ is a maximum.
Find and classify the stationary points of $y = x^3 - 6x^2 + 5$.
▶ Show solution
$3x^2 - 12x = 3x(x-4) = 0$, so $x = 0$ or $x = 4$.
At $x=0$: $y = 5$. At $x=4$: $y = 64 - 96 + 5 = -27$.
$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6x - 12$.
At $x=0$: $-12 < 0$, so $(0, 5)$ is a maximum.
At $x=4$: $12 > 0$, so $(4, -27)$ is a minimum.
Show that $y = x^3 + 3x$ has no stationary points.
▶ Show solution
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 + 3$
Since $x^2 \geqslant 0$ for all real $x$, we have $3x^2 + 3 \geqslant 3 > 0$.
The gradient is never zero, so there are no stationary points. The curve is always increasing.
Find the stationary points of $y = x^4 - 8x^2$.
▶ Show solution
$4x^3 - 16x = 4x\left(x^2 - 4\right) = 4x(x-2)(x+2) = 0$
$x = 0$, $2$ or $-2$.
At $x=0$: $y = 0$. At $x=\pm2$: $y = 16 - 32 = -16$.
Points: $(0, 0)$, $(2, -16)$ and $(-2, -16)$.
Classify the stationary points found in Question 7.
▶ Show solution
$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x^2 - 16$
At $x=0$: $-16 < 0$, so $(0,0)$ is a maximum.
At $x=2$: $48 - 16 = 32 > 0$, a minimum.
At $x=-2$: $48 - 16 = 32 > 0$, a minimum.
A "W" shape, with two equal minima either side of a local maximum at the origin.
The curve $y = x^3 + ax^2 + b$ has a stationary point at $(2, 1)$. Find $a$ and $b$.
▶ Show solution
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 + 2ax$, and this is zero at $x = 2$:
$12 + 4a = 0$, so $a = -3$.
The point $(2,1)$ is on the curve: $8 + a(4) + b = 1$
$8 - 12 + b = 1$, so $b = 5$.
Check: $y = x^3 - 3x^2 + 5$ gives $\tfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2-6x = 0$ at $x = 0, 2$ ✓ and $y(2) = 8-12+5 = 1$ ✓
The curve $C$ has equation $y = 2x^3 + 3x^2 - 36x + 5$.
(a) Find the coordinates of the stationary points. (b) Determine the nature of each. (c) Find the $y$-intercept and sketch $C$. (d) State the number of real solutions of $2x^3 + 3x^2 - 36x + 5 = 0$, with a reason.
▶ Show solution
(a) $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 + 6x - 36 = 6\left(x^2 + x - 6\right) = 6(x+3)(x-2)$
Zero when $x = -3$ or $x = 2$.
At $x = -3$: $y = 2(-27) + 3(9) + 108 + 5 = -54 + 27 + 108 + 5 = 86$
At $x = 2$: $y = 16 + 12 - 72 + 5 = -39$
Stationary points: $(-3, 86)$ and $(2, -39)$.
(b) $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x + 6$
At $x=-3$: $-36 + 6 = -30 < 0$, so $(-3, 86)$ is a maximum.
At $x=2$: $24 + 6 = 30 > 0$, so $(2, -39)$ is a minimum.
(c) $y$-intercept: $x = 0$ gives $y = 5$, so $(0, 5)$.
A positive cubic: rising from the bottom left, up to the maximum $(-3, 86)$, down through $(0, 5)$ to the minimum $(2, -39)$, then rising to the top right.
(d) The local maximum value is $86$, which is above the $x$-axis, and the local minimum value is $-39$, which is below it.
So the curve crosses the axis three times: once on the way up to the maximum, once falling from the maximum to the minimum, and once rising after the minimum.
The equation has three real solutions.
The general rule: a cubic has three real roots exactly when its two stationary values lie on opposite sides of the $x$-axis. If they were both positive or both negative, there would be only one root.