Notes
Two big theorems
- Mean Value Theorem (MVT): if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, there's a $c$ in $(a, b)$ with $\displaystyle f'(c) = \frac{f(b) - f(a)}{b - a}$. Some tangent slope equals the secant slope.
- Extreme Value Theorem: a continuous function on a closed interval $[a, b]$ has both an absolute max and an absolute min.
Increasing, decreasing and critical points
- $f' > 0$ means $f$ is increasing; $f' < 0$ means decreasing.
- A critical point is where $f' = 0$ or $f'$ is undefined (in the domain).
- First derivative test: $f'$ changes from $+$ to $-$ at a relative max, from $-$ to $+$ at a relative min.
Concavity
- $f'' > 0$: concave up ($f'$ increasing). $f'' < 0$: concave down.
- A point of inflection is where $f''$ changes sign (concavity changes).
- Second derivative test: if $f'(c) = 0$ and $f''(c) > 0$, relative min; if $f''(c) < 0$, relative max.
Absolute extrema on $[a, b]$ (candidates test)
Evaluate $f$ at the critical points and both endpoints. The largest value is the absolute max; the smallest is the absolute min.
Optimization
1. Write the quantity to maximize or minimize as a function of one variable. 2. Find its domain. 3. Find critical points and test them (and the endpoints). 4. Answer the question that was asked, with units.
Reading graphs of $f'$
When you're given the graph of $f'$: where it is above the axis, $f$ increases; where it crosses the axis from above to below, $f$ has a max; where $f'$ is increasing, $f$ is concave up.
Topics in this unit
Each topic has its own short notes and practice.
- 5.1Using the Mean Value Theorem
- 5.2Extreme Value Theorem, Global Versus Local Extrema, and Critical Points
- 5.3Determining Intervals on Which a Function Is Increasing or Decreasing
- 5.4Using the First Derivative Test to Determine Relative (Local) Extrema
- 5.5Using the Candidates Test to Determine Absolute (Global) Extrema
- 5.6Determining Concavity of Functions over Their Domains
- 5.7Using the Second Derivative Test to Determine Extrema
- 5.8Sketching Graphs of Functions and Their Derivatives
- 5.9Connecting a Function, Its First Derivative, and Its Second Derivative
- 5.10Introduction to Optimization Problems
- 5.11Solving Optimization Problems
- 5.12Exploring Behaviors of Implicit Relations
Practice
Secant slope $\displaystyle \frac{9 - 1}{2} = 4$. $f'(c) = 2c = 4 \Rightarrow c = 2$.
$f'(x) = 3x^2 - 12 = 3(x - 2)(x + 2) < 0$ for $-2 < x < 2$.
$f''(x) = 6x - 12$ changes sign at $x = 2$.
Second derivative test: $f'' < 0$ means concave down, so a max.
Critical point $x = 1$ gives $f(1) = -2$. Endpoints: $f(0) = 0$, $f(3) = 27 - 9 = 18$.
$f'$ changes from negative to positive, so $f$ changes from decreasing to increasing.
$A = x(200 - x)$, max at $x = 100$: a 100 by 100 square with area 10,000 m².
$f'$ increasing means $f'' > 0$, which means concave up.
Free response
Write your answer first, then open the scoring guide.
Free-response question
Let $f(x) = xe^{-x}$ for all real $x$.
(a) Find $f'(x)$ and the $x$-coordinate of every critical point.
Show the answer and scoring guide
$f'(x) = e^{-x} - xe^{-x} = (1 - x)e^{-x}$. Since $e^{-x} > 0$, the only critical point is $x = 1$.
Check off the points you earned:
(b) Does $f$ have a relative max or min at $x = 1$? Justify.
Show the answer and scoring guide
A relative max: $f'(x) = (1 - x)e^{-x}$ changes from positive to negative at $x = 1$.
Check off the points you earned:
(c) Find the $x$-coordinate of the point of inflection. Justify.
Show the answer and scoring guide
$f''(x) = -e^{-x} - (1 - x)e^{-x} = (x - 2)e^{-x}$. It changes sign from negative to positive at $x = 2$, so $x = 2$ is a point of inflection.
Check off the points you earned: