Unit 5: Analytical Applications of Differentiation

17 terms. Say what each one means, then open the card to check.

• Mean Value Theorem

  If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then $f'(c) = \frac{f(b) - f(a)}{b - a}$ for at least one $c$ strictly between $a$ and $b$.

  Example. A hiker whose elevation changes smoothly and who climbs 600 meters in 2 hours is climbing at exactly 300 meters per hour at some moment.

• Conditions of the Mean Value Theorem

  Continuity on the closed interval and differentiability on the open interval. If either fails, the conclusion may fail.

  Example. $f(x) = |x|$ on $[-1, 1]$ has an average rate of 0, but $f'$ is never 0, because $f$ has a corner at $x = 0$.

• Extreme Value Theorem

  A function that is continuous on a closed interval has both an absolute maximum and an absolute minimum on that interval.

  Example. $f(x) = \frac{1}{x}$ on $(0, 1]$ has no maximum: the interval is not closed, so the theorem does not apply.

• Critical point

  An input $c$ in the domain of $f$, not an endpoint, where $f'(c) = 0$ or $f'(c)$ does not exist.

  Example. $f(x) = x^{2/3}$ has a critical point at $x = 0$, where $f'$ is undefined and the graph has a cusp.

• Relative extremum

  A relative (local) maximum or minimum: a value of f that is the largest or smallest among the values at nearby inputs.

  Example. A hill on a graph at $x = 2$ is a relative maximum even if the graph climbs higher later on.

• Absolute extremum

  The largest or smallest value of f on its whole domain or on a stated interval.

  Example. On $[0, 10]$, a profit function might have its absolute maximum at the endpoint $x = 10$, where $f'$ need not be 0.

• First derivative test

  At a critical point, a change in the sign of $f'$ from positive to negative gives a relative maximum; from negative to positive, a relative minimum; no change, neither.

  Example. If $f'$ is negative just left of $x = 1$ and positive just right of it, $f$ has a relative minimum at $x = 1$.

• Candidates test

  To find absolute extrema of a continuous function on a closed interval, evaluate f at every critical point and both endpoints, then compare the values.

  Example. For $k$ on $[0, 4]$ with one critical point at $x = 3$, compare $k(0)$, $k(3)$ and $k(4)$; the largest is the absolute maximum.

• Second derivative test

  If $f'(c) = 0$ and $f''(c) < 0$, f has a relative maximum at $c$; if $f''(c) > 0$, a relative minimum. If $f''(c) = 0$, the test gives no answer.

  Example. $f'(1) = 0$ and $f''(1) = 6$ mean the graph is concave up at a flat point, so $f$ has a relative minimum at $x = 1$.

• Increasing and decreasing

  f is increasing on an interval where $f' > 0$ and decreasing where $f' < 0$.

  Example. If $f'(x) = e^x$, then $f' > 0$ everywhere, so $f$ is increasing on every interval.

• Concavity

  The graph of f is concave up where $f'$ is increasing ($f'' > 0$) and concave down where $f'$ is decreasing ($f'' < 0$).

  Example. $y = x^2$ is concave up everywhere: its slopes increase from left to right.

• Point of inflection

  A point on the graph where the concavity changes, which happens where $f''$ changes sign. $f''$ being 0 there is not enough on its own.

  Example. $f(x) = x^4$ has $f''(0) = 0$, but $f'' \geq 0$ on both sides, so there is no inflection point at $x = 0$.

• Reading f from the graph of its derivative

  Where the graph of $f'$ is above the axis, f increases; where it crosses the axis, f may have an extremum; where it rises or falls, f is concave up or down.

  Example. If the graph of $f'$ has a peak at $x = 4$, then $f'$ switches from increasing to decreasing there, so $f$ has an inflection point at $x = 4$.

• Objective function

  In an optimization problem, the quantity to be made as large or as small as possible, written as a function of one variable.

  Example. To build the cheapest can of a given volume, the objective function is the cost of the metal, written in terms of the radius.

• Constraint

  An equation that a problem’s quantities must satisfy. Use it to remove all but one variable from the objective function.

  Example. A fixed volume $\pi r^2 h = 500$ lets you write $h = \frac{500}{\pi r^2}$ and turn surface area into a function of $r$ alone.

• Feasible domain

  The inputs that make physical sense in an optimization problem. Critical points outside it are thrown out, and its endpoints must be checked.

  Example. A side length must be positive and can be no longer than the material allows, so a box’s side might be limited to 0 < x < 6.

• Single critical point argument

  If a continuous function on an interval has only one critical point, and it is a relative minimum (or maximum), then it is the absolute minimum (or maximum) on that interval.

  Example. If cost has one critical point, at $x = 12$, with $C'$ going from negative to positive, then $x = 12$ gives the lowest cost.