Notes
Increasing and decreasing, extrema
- $f' > 0$ means increasing, $f' < 0$ means decreasing. Critical numbers are where $f' = 0$ or is undefined.
- First derivative test: $+$ to $-$ is a local max, $-$ to $+$ is a local min.
- Second derivative test: when $f'(c) = 0$, $f''(c) > 0$ is a local min and $f''(c) < 0$ is a local max.
Concavity
- $f'' > 0$ is concave up, $f'' < 0$ is concave down. An inflection point is where concavity changes.
- $f''(c) = 0$ alone does not guarantee an inflection point. For $f(x) = x^4$, $f''(0) = 0$ but concavity never changes.
Absolute extrema and optimization
- On $[a, b]$, compare $f$ at critical numbers and at the endpoints.
- Optimization: write the quantity as a function of one variable using the constraint, then find its maximum or minimum.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$f' = 3x^2 - 12 = 3(x - 2)(x + 2)$.
Falling then rising means a valley.
$6x - 18 > 0$ when $x > 3$.
Critical number is $x = 3$. Values: $f(0) = 10$, $f(3) = 1$, $f(5) = 5$.
$P = x(12 - x)$ is maximized at $x = 6$, giving 36.
$f'' = 12x^2$ is positive on both sides of 0, so concavity does not change.
Free response
Write your answer first, then open the scoring guide.
Free-response question
Let $f(x) = x^3 - 6x^2 + 9x + 1$ on the interval $[0, 4]$.
(a) Find the critical numbers and the intervals where $f$ is increasing and decreasing.
Show the answer and scoring guide
$f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$. The critical numbers are $x = 1$ and $x = 3$. $f' > 0$ for $x < 1$ and $x > 3$ (increasing), and $f' < 0$ for $1 < x < 3$ (decreasing).
Check off the points you earned:
(b) Classify each critical number as a local max or min and give the value.
Show the answer and scoring guide
The sign of $f'$ goes $+, -, +$. So $x = 1$ is a local max with $f(1) = 5$, and $x = 3$ is a local min with $f(3) = 1$.
Check off the points you earned:
(c) Find the absolute maximum and minimum values of $f$ on $[0, 4]$.
Show the answer and scoring guide
$f(0) = 1$, $f(1) = 5$, $f(3) = 1$, $f(4) = 5$. The absolute maximum is 5 and the absolute minimum is 1.
Check off the points you earned: