Monotonicity and Extrema
$$f'(x) > 0 \Rightarrow \text{increasing}$$
$$f'(x) < 0 \Rightarrow \text{decreasing}$$
The second derivative $f''(x)$ tells us how the graph of a function bends. Understanding concavity helps us identify inflection points and determine whether critical points are maxima or minima.
Concavity of the graph:
Inflection points:
An inflection point occurs where the graph changes concavity — from concave up to concave down or vice versa. This usually happens where $f''(x)$ changes sign.
The Second Derivative Test for maxima and minima:
If a point $x = c$ is a critical point (meaning $f'(c) = 0$), then the second derivative helps determine whether it is a peak or a valley.
Taken together, concavity and the second derivative test provide a complete picture of how the function behaves: where it bends, where it reaches peaks and valleys, and where the direction of bending changes.
How to solve optimization problems:
These problems usually involve finding the best (maximum or minimum) value under certain conditions, such as fixed perimeter, fixed area or volume, limited materials, or being restricted to a given path or shape.
Where is the point of inflection of the graph $y = (x+2)^{3/5} - 6x$?
At a point of inflection, $y''$ changes sign (or is undefined and changes sign).
$y''$ is undefined at $x = -2$. Since the sign of $y''$ changes there, $x = -2$ is an inflection point.
Point of inflection: $(-2,\ 12)$
Find the point of inflection of the curve $y = \dfrac{6x^2 - x^3 + 5}{7}$.
Point of inflection: $(2,\ 3)$
Find the absolute extrema of $f(x) = x^3 - 3x^2 - 24x + 2$ on the intervals $[-3,\,1]$ and $[-3,\,8]$.
Find critical points:
On $[-3,\,1]$ (only $x=-2$ is inside):
| $x$ | $f(x)$ |
|---|---|
| $-3$ | $20$ |
| $-2$ | $\mathbf{30}$ (max) |
| $1$ | $\mathbf{-24}$ (min) |
On $[-3,\,8]$ (both $x=-2$ and $x=4$ are inside):
| $x$ | $f(x)$ |
|---|---|
| $-3$ | $20$ |
| $-2$ | $30$ |
| $4$ | $\mathbf{-78}$ (min) |
| $8$ | $\mathbf{130}$ (max) |
Find the critical points of $f(x) = \dfrac{x^2}{x-1}$.
$f'(x) = 0$ when $x = 0$ or $x = 2$. ($x=1$ is not in the domain.)
Critical points: $x = 0$ and $x = 2$
A closed cylindrical tank is to be made using steel plates with a total surface area of 115 m². What is the greatest volume possible?
Surface area: $2\pi r h + 2\pi r^2 = 115$, so $h = \dfrac{115/(2\pi) - r^2}{r}$.
If the equal sides of an isosceles triangle are each of length $x$, what length of the third side $y$ will provide the maximum area?
Using the apex half-angle $\theta$: area $= x^2 \sin\theta\cos\theta = \frac{x^2}{2}\sin 2\theta$.
Maximum when $\sin 2\theta = 1 \Rightarrow \theta = 45°$.
The third side for maximum area is $y = x\sqrt{2}$ (right isosceles triangle).
A 20 ft long water trough has ends in the form of isosceles triangles with sides that are 4 ft long. Determine the maximum volume of the trough.
Let $x$ = base width, $h$ = height of the triangular end.