Cubic Graphs, Optimisation and Rates of Change

Sketching a cubic

For $f(x) = ax^3 + bx^2 + cx + d$ you need: the intercepts, the stationary points (where $f'(x) = 0$) and the point of inflection (where $f''(x) = 0$, i.e. $x = -\dfrac{b}{3a}$).

Worked example: $f(x) = x^3 - 6x^2 + 9x$

Intercepts. $f(0) = 0$. Factorising, $x(x^2-6x+9) = x(x-3)^2$, so $x = 0$ or $x = 3$ (a touch point).

Stationary points. $f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$, so $x = 1$ or $x = 3$.

• $f(1) = 1 - 6 + 9 = 4$ — local maximum at $(1; 4)$.
• $f(3) = 27 - 54 + 27 = 0$ — local minimum at $(3; 0)$.

Inflection. $f''(x) = 6x - 12 = 0$ at $x = 2$, and $f(2) = 8 - 24 + 18 = 2$, so $(2; 2)$.

Optimisation

To maximise or minimise a quantity: write it as a function of one variable using the constraint, differentiate, set the derivative to zero, and solve.

Worked example: a vegetable garden

A school has 40 m of fencing for a rectangular garden against an existing wall (no fence needed along the wall). If the two equal sides are $w$ metres, then the third side is $40 - 2w$ and

So the sides are 10 m, 10 m and 20 m, giving a maximum area of $200\ \text{m}^2$.

Rates of change

If $s(t)$ is distance then $s'(t)$ is velocity and $s''(t)$ is acceleration. The derivative always answers "how fast is this quantity changing right now?"