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Flashcards · AP® Calculus BC · Unit 5

Unit 5: Analytical Applications of Differentiation flashcards

74 cards on topics 5.1 to 5.12. Pick a topic or study the whole unit.

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5.1 Using the Mean Value Theorem

What does the Mean Value Theorem state?
If f is continuous on [a, b] and differentiable on (a, b), then at least one c in (a, b) has f′(c) = (f(b) − f(a))/(b − a): a tangent inside is parallel to the secant through the endpoints.
Where must continuity and differentiability hold for the Mean Value Theorem?
Continuity on the whole closed interval [a, b], endpoints included; differentiability only inside (a, b). A corner, cusp, vertical tangent, jump, hole or asymptote inside the interval breaks a condition.
Find the value of c guaranteed by the Mean Value Theorem for f(x) = x² + 2x on [0, 2].
Average rate (f(2) − f(0))/2 = (8 − 0)/2 = 4. f′(c) = 2c + 2 = 4 gives c = 1, which is inside (0, 2).
A differentiable function has g(2) = 7 and g(6) = −1. What must be true?
By the Mean Value Theorem, g′(c) = (−1 − 7)/(6 − 2) = −2 for some c in (2, 6).
How does the Mean Value Theorem differ from the Intermediate Value Theorem?
The Intermediate Value Theorem guarantees values of f (needs continuity). The Mean Value Theorem guarantees a value of f′ equal to the average rate (needs continuity and differentiability).
What does the Mean Value Theorem mean in a motion context?
At some moment in the interval, the instantaneous velocity equals the average velocity over the interval.
What mistakes are common with the Mean Value Theorem?
Taking c as the midpoint; averaging f′(a) and f′(b); keeping a c at an endpoint or outside the interval; omitting the conditions or the theorem's name; saying "no such c" when a condition fails (only the guarantee is lost).

5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points

What does the Extreme Value Theorem state?
If f is continuous on the closed interval [a, b], then f has at least one absolute maximum value and at least one absolute minimum value on [a, b]. It does not say where.
What is a critical point?
A value c in the domain of f where f′(c) = 0 or f′(c) does not exist. Points where f itself is undefined are not critical points.
Is every critical point a local extremum?
No. Every local extremum is at a critical point, but not conversely. x³ has f′(0) = 0 with no extremum, and ∛x has a vertical tangent at 0 with no extremum.
What is the difference between a global and a local extremum?
Global (absolute): highest or lowest value on the whole interval, possibly at an endpoint. Local (relative): highest or lowest compared with points on an open interval around c, so an endpoint is not a local extremum.
Is x = 5 a critical point of f(x) = |x − 5|?
Yes. f is defined at 5 but has a corner there, so f′(5) does not exist.
How do you find all critical points efficiently?
Write f′ as one fraction. Zeros of the top give f′ = 0; zeros of the bottom give f′ undefined (check f is defined there). Do not stop at f′ = 0.
What happens to the Extreme Value Theorem on an open interval or across a break?
The guarantee is lost. A failed condition means "no guarantee", not "no extremum".

5.3 Determining Intervals on Which a Function Is Increasing or Decreasing

How does f′ tell you where f is increasing or decreasing?
If f′(x) > 0 on an interval, f is increasing there; if f′(x) is negative on an interval, f is decreasing there. Answers are intervals.
Why does the sign of f′ decide increasing or decreasing?
By the Mean Value Theorem, f(x₂) − f(x₁) = f′(c)(x₂ − x₁), so the sign of f′ decides the sign of the change.
f′(x) = (x − 3)². Where is f increasing?
Everywhere: f′ > 0 except at x = 3, where f′ = 0. An isolated zero with the same sign on both sides does not break the interval, so f is increasing on (−∞, ∞).
q(x) = x² − 8 ln x for x > 0. Where is q decreasing?
q′(x) = 2x − 8/x = 2(x² − 4)/x, which is negative on (0, 2). So q is decreasing on (0, 2) and increasing on (2, ∞).
The graph of g′ is falling but above the x-axis. Is g increasing or decreasing?
Increasing, because g′ > 0. The direction of g′ is about concavity, not about whether g rises or falls.
What mistakes are common with increasing and decreasing intervals?
Reading the direction of f′ instead of its sign; forgetting domain gaps (x + 4/x is undefined at 0); assuming signs alternate past a squared factor; justifying with the graph of f; writing "it" instead of naming f′.

5.4 Using the First Derivative Test to Determine Relative (Local) Extrema

What is the First Derivative Test?
At a critical point c where f is continuous: f′ changes + to − gives a relative maximum; − to + gives a relative minimum; no sign change gives neither.
On a graph of f′, what shows a relative maximum or minimum of f?
f′ crossing the axis downwards: relative maximum. Crossing upwards: relative minimum. Touching the axis and turning back: neither. A peak of f′ away from the axis is not an extremum of f.
m(x) = x³ − 3x. Classify its critical points.
m′ = 3(x − 1)(x + 1). At x = −1, m′ changes + to −: relative maximum m(−1) = 2. At x = 1, − to +: relative minimum m(1) = −2.
f′(x) = (x − 2)³. Does f have a relative extremum at x = 2?
Yes, a relative minimum: f′ is negative for x less than 2 and positive for x greater than 2.
Why is "f′(c) = 0, so f has a maximum" not a valid justification?
It does not show the type. Say how the sign of f′ changes at c, e.g. "f′ changes from positive to negative at x = c".
What mistakes are common with the First Derivative Test?
Missing critical points where f′ is undefined (e.g. cusps); substituting into f′ instead of f for the value; giving the x-value when the value is asked; applying the test at a jump or hole.

5.5 Using the Candidates Test to Determine Absolute (Global) Extrema

What is the Candidates Test for absolute extrema?
For f continuous on [a, b]: find critical points strictly inside (a, b), evaluate f at each and at a and b, then the largest value is the absolute maximum and the smallest is the absolute minimum.
Why are critical points and endpoints the only candidates for absolute extrema?
An interior absolute extremum is also a local extremum, so it sits at a critical point; otherwise it sits at an endpoint. The Extreme Value Theorem guarantees both exist on a closed interval.
Find the absolute extrema of f(x) = x³ − 3x on [0, 2].
f′ = 3x² − 3 = 0 at x = 1. Values: f(0) = 0, f(1) = −2, f(2) = 2. Absolute max 2 at x = 2; absolute min −2 at x = 1.
f(x) = 2x³ − 9x² + 12x on [0, 3]. What is the absolute maximum value?
Critical points x = 1, 2. f(0) = 0, f(1) = 5, f(2) = 4, f(3) = 9. The absolute maximum is 9 at the endpoint x = 3, even though x = 1 is a local maximum.
Does x² have an absolute maximum on the open interval (−1, 2)?
No. Values approach 4 near x = 2 but never reach it. The Candidates Test needs a closed interval.
How should you state an absolute extremum?
"The absolute maximum value of f on [a, b] is [value], at x = [value]." Use f (not f′) values, and keep full precision until you compare.

5.6 Determining Concavity of Functions over Their Domains

What does concave up or concave down mean?
Concave up: slopes increase (f′ increasing, or f″ > 0), and the graph bends like a cup. Concave down: slopes decrease (f′ decreasing, or f″ negative), bending like a cap.
What is a point of inflection and how do you find one?
A point on the graph where concavity changes. Candidates are where f″ = 0 or f″ is undefined; confirm f″ changes sign there and f(c) exists.
Does f″(c) = 0 guarantee a point of inflection?
No. x⁴ has f″(0) = 0 but is concave up on both sides, so there is no inflection point.
Where is f(x) = x³ − 3x² + 1 concave down, and where is its inflection point?
f″ = 6x − 6 is negative for x less than 1, so concave down on (−∞, 1). Inflection point (1, −1).
On a graph of f′, what shows an inflection point of f?
A local maximum or minimum of the f′ graph. Where f′ crosses the x-axis gives a local extremum of f, not an inflection point.
How do you describe f′ > 0 and f″ less than 0 in context?
"Increasing at a decreasing rate". Avoid vague wording like "slowing down".

5.7 Using the Second Derivative Test to Determine Extrema

What is the Second Derivative Test?
If f′(c) = 0 and f″(c) > 0, f has a relative minimum at c (cup). If f′(c) = 0 and f″(c) is negative, a relative maximum (cap). If f″(c) = 0 the test is inconclusive.
Let h(x) = x³ − 6x² + 2. Classify its critical points with the Second Derivative Test.
h′ = 3x(x − 4), h″ = 6x − 12. h″(0) = −12: relative maximum at x = 0 (h = 2). h″(4) = 12: relative minimum at x = 4 (h = −30).
f′(2) = 0 and f″(2) = 0. What can you conclude?
Nothing yet. Use the First Derivative Test: check the sign of f′ on each side. k(x) = x⁴ + 2 has k″(0) = 0 but a relative minimum at 0.
What is the one-critical-point rule?
If f is continuous on an interval with exactly one critical point c, and c is a relative minimum (maximum), then f(c) is the absolute minimum (maximum) on that interval.
What is a model justification using the Second Derivative Test?
"f has a relative minimum at x = 4 because f′(4) = 0 and f″(4) > 0." Include f′(c) = 0, not just the sign of f″.
When does the Second Derivative Test not apply?
When f′(c) does not exist (a cusp or corner) or f″(c) = 0. Use the First Derivative Test instead.

5.8 Sketching Graphs of Functions and Their Derivatives

How do you sketch f using a sign chart?
Find where f′ and f″ are zero or undefined, put them on one number line, test signs in each interval, calculate f at each marked point, then sketch with the right rise and bend.
What shape does f have for each sign combination of f′ and f″?
f′ +, f″ +: rising, getting steeper. f′ +, f″ −: rising, flattening. f′ −, f″ +: falling, flattening. f′ −, f″ −: falling, getting steeper.
Let r(x) = x³ − 12x. Give the relative extrema and the inflection point.
r′ = 3(x − 2)(x + 2), r″ = 6x. Relative maximum (−2, 16), relative minimum (2, −16), inflection point (0, 0).
The graph of f′ touches the x-axis at x = 5 without crossing it. Is there an extremum of f at 5?
No: f′ keeps its sign. But f′ has a peak or trough there, so f has an inflection point with a horizontal tangent.
What does a peak or trough on the graph of f′ mean for f?
A point of inflection of f, not an extremum.
What must a sketch of f show?
The correct bend between key points (not straight-line joins) and labelled coordinates of every extremum and inflection point.

5.9 Connecting a Function, Its First Derivative, and Its Second Derivative

What do the sign and direction of the f′ graph tell you about f?
Sign of f′ (above or below the axis): f increasing or decreasing. Direction of f′ (rising or falling): f concave up or down.
What does the sign of f″ tell you, and what does it not tell you?
It tells you whether f is concave up or down and whether f′ is increasing or decreasing. It does not tell you whether f is rising or falling.
f′(x) = x³ − 6x² = x²(x − 6). Where does f have a relative extremum?
Only at x = 6, a minimum: f′ changes from negative to positive. At x = 0, f′ touches 0 without changing sign.
For f′(x) = x³ − 6x², where are the inflection points of f?
f″(x) = 3x² − 12x = 3x(x − 4), which changes sign at x = 0 and x = 4.
How do you match graphs of f, f′ and f″?
Zeros of one curve line up with turning points of the curve one level up. Check signs too, then build the chain f → f′ → f″.
Can a graph of f′ give you the values of f?
No. It only shows how f changes. Reading f(c) from the height of the f′ graph is a common mistake.

5.10 Introduction to Optimization Problems

What is the skeleton of every optimization problem?
A quantity to optimise, a constraint, and an interval. Use the constraint to write the quantity as a function of one variable, then differentiate.
What are the eight steps of the optimization method?
Name the quantity → diagram and letters → formula → use the constraint → interval → critical points → prove it is the maximum or minimum → answer what was asked.
Two numbers add to 20. What is the largest possible product?
P = x(20 − x), P′ = 20 − 2x = 0 at x = 10, P″ = −2 negative, and it is the only critical point, so the maximum product is 100.
Find the maximum and minimum of f(x) = x³ − 12x on [0, 3].
f′ = 3x² − 12 = 0 at x = 2. f(0) = 0, f(2) = −16, f(3) = −9. Maximum 0 at x = 0; minimum −16 at x = 2.
On an open interval, g′(c) = 0 and g″(c) is negative. Is g(c) the absolute maximum?
Not yet. That only shows a relative maximum. You also need c to be the only critical point on the interval, or a comparison of values.
What mistakes are common in setting up optimization?
Differentiating a formula that still has two variables; stopping at f′ = 0 without proof; forgetting endpoints; keeping a critical point outside the interval, such as a negative length.

5.11 Solving Optimization Problems

How do you justify an absolute extremum in an optimization problem?
Closed domain: Candidates Test. Open domain with one critical point: a sign change of f′ or the sign of f″(c), plus "only critical point". f″ of one sign on the whole domain also works.
On [0, 10], f has one critical point at x = 4 with f(4) = 7, and f(0) = 2, f(10) = 9. What is the absolute maximum value?
9, at x = 10. The endpoint wins.
On x > 0, g has only one critical point, at x = 5, and g″(5) > 0. What can you conclude?
g(5) is the absolute minimum of g on x > 0.
R(t) is a rate in litres per minute with its maximum at t = 12. What happens at t = 12?
The liquid flows fastest. It is not when the most liquid has flowed in total: the maximum of a rate is the fastest rate.
How should an optimization answer be given?
Give exactly what is asked (location, value or both), name the quantity, give units and say when or where. On calculator questions give three decimal places and do not round early.
Is f″(c) less than 0 alone enough for an absolute maximum?
No. It gives a relative maximum; you also need the one-critical-point reason or a comparison of values.

5.12 Exploring Behaviors of Implicit Relations

How do you find critical points of an implicitly defined curve?
Write dy/dx = N/D. Set N = 0 (horizontal tangent) or D = 0 (vertical tangent), then solve together with the curve's equation, and check the other part of the fraction at each point.
What does N = 0 and D = 0 at a point on an implicit curve mean?
dy/dx is 0/0: the formula decides nothing, and you must investigate another way. It is not automatically a horizontal tangent.
dy/dx = (x − 1)/(y + 3). Where could the tangent be vertical?
Where y = −3 on the curve, provided x ≠ 1 there.
For y² = x² − 4x + 13, 2(y′)² + 2y·y″ = 2. Classify the horizontal tangent at (2, 3).
y′ = 0 there, so 6y″ = 2 and y″ = 1/3 > 0: y has a relative minimum at x = 2.
dy/dx = x/y. In which quadrants is y decreasing?
II and IV, where x and y have opposite signs so dy/dx is negative.
What is a shortcut for y″ at a horizontal tangent of an implicit curve?
Every term containing y′ is 0 there. You still need both coordinates of the point to evaluate y″.

Every card in this unit is shared with the AP® Calculus AB flashcards, because the two courses share these topics.

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