Calculus 1C — Applications of the Derivative

Maxima, minima & curve shape

continues from lesson 2 — values defined earlier in the course stay live here

ELI5: a hilltop or a valley-bottom is flat for an instant — slope zero. So candidates for max/min are where f′(x) = 0. To tell which is which, look at the second derivative (the bend): f″ > 0 means the curve cups up (a smile → a minimum); f″ < 0 means it cups down (a frown → a maximum).

Take f(x) = x³ − 3x. Then f′(x) = 3x² − 3 = 0 at x = ±1. Since f″ = 6x: f″(−1) = −6 (a frown → local max) and f″(1) = +6 (a smile → local min).

f_{hv}\left(x\right) = x^{3} - 3 \cdot x

hill_{x} = -1

local max from 3x² − 3 = 0

valley_{x} = 1

local min

✓ pass

\mathrm{f_{hv}}\left(hill_{x}\right) > \mathrm{f_{hv}}\left(valley_{x}\right)

the hilltop at x = −1 sits above the valley at x = 1

f_hv(x)

the hill (x = −1) and the valley (x = 1) of x³ − 3x

Real-world hook: maxima and minima are where you read off peak power, maximum range, lowest cost — and the inflection (where f″ flips sign) is where growth stops accelerating and starts to ease off.

Try it yourself: f(x) = x³ − 6x² + 9x has two critical points. Find the x of its local maximum.

f_{cm}\left(x\right) = x^{3} - 6 \cdot x^{2} + 9 \cdot x

xmax = 1

✏️ Your turn: replace 0 with the x where x³ − 6x² + 9x has a local MAX. The check confirms it beats both neighbours — so it never reveals the answer.

✓ pass

f_{cm}(xmax) >= f_{cm}(xmax - 0.3) and f_{cm}(xmax) >= f_{cm}(xmax + 0.3)

green when your point really is higher than its neighbours