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B. Derivatives and Their Applications · Curve Sketching: Connecting f, f′ and f″

Reading f from Information About f′ and f″

B1.4 B1.1About 70 minutes
Learning goals
  • I can describe the key features of a function from information about its first and second derivatives.
  • I can sketch two or more possible graphs that fit given derivative information.
  • I can explain why derivative information leads to infinitely many possible graphs, and what extra information pins one down.

Words to know

family of functions
A set of functions that share a property, such as all functions with derivative 2x (namely x² + C).
arbitrary constant C
The unknown constant in f(x) + C; one point on the graph determines it.

Working backwards

So far you have started with f and found its derivatives. Now flip it around. Suppose a weather station gives you only the rate at which the temperature changed through a day, not the temperature itself. What can you say about the temperature graph? You can describe its shape exactly, but not its height unless you are told one actual temperature.

That is the big idea of this lesson: information about f′ and f″ fixes the shape of f, but many graphs share that shape.

Why infinitely many graphs?

  1. Vertical shifts. If f fits the conditions, so does f(x) + C for every constant C, because (f + C)′ = f′. There are infinitely many values of C.
  2. Sign information is loose. If you only know the signs of f′ and f″ on intervals, you do not know how steep the graph is. Stretching it vertically by any positive factor, or bending it differently while keeping the same signs, still fits.

To pin down one graph you need the exact formula for f′ and one point on the graph.

Turning derivative information into a sketch

  1. Make one sign chart with rows for f′ and f″.
  2. Combine the rows into a shape for each interval: rising/falling and ∪/∩ (for example, "rising and concave down").
  3. Mark the special points: local max/min where f′ changes sign; inflection points where f″ changes sign.
  4. Plot any given points, then draw a smooth curve with the right shape on each interval.
  5. To show that many graphs are possible, draw a second graph with the same shapes but different heights or steepness.
Worked example

Example 1: Signs only

Sketch two possible graphs of a continuous function f with: f′(x) < 0 for x < −1 and x > 3; f′(x) > 0 for −1 < x < 3; f″(x) > 0 for x < 1; f″(x) < 0 for x > 1.

  1. f′: − then + at x = −1 (local minimum); + then − at x = 3 (local maximum).
  2. f″: + then − at x = 1 (inflection point, concave up to concave down).
  3. Combine: x < −1, falling and ∪; −1 < x < 1, rising and ∪; 1 < x < 3, rising and ∩; x > 3, falling and ∩.
  4. Choose heights: graph A puts the minimum at (−1, −2) and the maximum at (3, 6); graph B uses (−1, 2) and (3, 6). Both have the right shape on every interval.
  5. A cubic with negative leading coefficient fits this pattern, but so do many non-polynomial curves.

See the two graphs below. Infinitely many graphs fit, because only signs are given: any vertical shift or positive vertical stretch of a valid graph is also valid.

graph Axy−2−11234−4−22468min (−1, −2)IP (1, 2)max (3, 6)graph Bxy−2−11234−4−22468min (−1, 2)IP (1, 4)max (3, 6)
Both graphs have exactly the sign pattern of Example 1: decreasing, then increasing, then decreasing, with the concavity switching at x = 1. Graph B is graph A squashed vertically and moved up. Infinitely many other graphs fit too.
Worked example

Example 2: A curriculum investigation

Let g(x) = (x − 1)(x − 3) and suppose g(x) = f′(x). (a) Sketch f if f(0) = 0. (b) Find f″(x). (c) How does the graph change if f(0) = 2?

  1. Expand: f′(x) = x² − 4x + 3. A function with this derivative is f(x) = (1/3)x³ − 2x² + 3x + C. (Check by differentiating: x² − 4x + 3 ✓.)
  2. (a) f(0) = C = 0, so f(x) = (1/3)x³ − 2x² + 3x.
  3. f′ = (x − 1)(x − 3): + for x < 1, − for 1 < x < 3, + for x > 3. Local max at x = 1: f(1) = 1/3 − 2 + 3 = 4/3. Local min at x = 3: f(3) = 9 − 18 + 9 = 0.
  4. (b) f″(x) = 2x − 4, zero at x = 2: concave down for x < 2, up for x > 2. Inflection point: f(2) = 8/3 − 8 + 6 = 2/3.
  5. (c) If f(0) = 2, then C = 2. Every point moves up 2: max (1, 10/3), min (3, 2), inflection point (2, 8/3). f′ and f″ do not change.

(a) Cubic through (0, 0) with max (1, 4/3), min (3, 0), IP (2, 2/3). (b) f″(x) = 2x − 4. (c) The same graph translated up 2 units.

f′(x) = (x − 1)(x − 3)xy1234−11234f(0) = 0f(0) = 2(1, 4/3)
Two members of the family with derivative (x − 1)(x − 3). The solid curve has its maximum at (1, 4/3) and its minimum at (3, 0). Changing f(0) from 0 to 2 lifts the whole graph by 2 units; the x-coordinates of the turning points and the inflection point do not change.
Worked example

Example 3: Reading f from the graph of f′

The graph of f′(x) = ¼(x + 2)²(x − 1) is shown below. Describe f: intervals of increase and decrease, extrema, concavity and inflection points.

  1. f′ is negative for x < −2 and for −2 < x < 1 (it only touches 0 at −2), and positive for x > 1.
  2. So f decreases on (−∞, 1) and increases on (1, ∞). Local minimum at x = 1. At x = −2 the tangent is horizontal but there is no turn.
  3. Concavity comes from whether f′ is rising or falling. f′ rises until x = −2, falls from −2 to 0 (its lowest point), and rises after 0.
  4. So f is concave up on (−∞, −2), concave down on (−2, 0) and concave up on (0, ∞). Inflection points at x = −2 (horizontal) and x = 0.
  5. Algebra check: f″(x) = ¼[2(x + 2)(x − 1) + (x + 2)²] = ¼(x + 2)(3x), zero at −2 and 0 with sign changes. ✓

Decreasing on (−∞, 1), increasing on (1, ∞); local min at x = 1; inflection points at x = −2 (horizontal) and x = 0; concave up, down, up.

given: y = f′(x)xy−4−3−2−112−3−2−112touches 0crosses 0lowest pointone possible fxy−4−3−2−112−11234flat IPIPmin
Left: the graph of f′ touches the axis at x = −2 (no sign change) and crosses at x = 1 (− to +). Its lowest point is at x = 0. Right: so f has a horizontal inflection point at x = −2, a minimum at x = 1, and another inflection point at x = 0.
Worked example

Example 4: Describing a polynomial from its derivatives

A polynomial f has degree 4 and positive leading coefficient. Its derivative has zeros only at x = −3, 0 and 2, and they are all simple (each factor appears once). Describe f.

  1. f′ is a cubic with positive leading coefficient: f′ = a(x + 3)x(x − 2), a > 0.
  2. Signs of f′: −, +, −, + . So local min at −3, local max at 0, local min at 2.
  3. f″ is a quadratic with two zeros, one between −3 and 0 and one between 0 and 2 (each turning point of f′ lies between two of its zeros). So f has two inflection points.
  4. End behaviour: f → ∞ as x → ±∞ (a W shape).

f is W-shaped: minimums at x = −3 and 2, a maximum at 0, and two inflection points, one in (−3, 0) and one in (0, 2). Without f-values or the value of a, there are infinitely many such graphs.

Just for fun

Calculus is one of the few subjects where adding "+ C" to your answer means "I know exactly what you look like, I just don't know how high up you are."

Hands-on activity

Describe it, draw it

You need: a partner, two pencils, two sheets of paper, and a book to hide your drawing.

  1. Secretly sketch a smooth curve with at least one maximum, one minimum and one inflection point.
  2. Without showing it, describe it to your partner using only derivative language: "f′ is positive for x < 1, negative from 1 to 4..." and "f″ is negative for x < 2.5...". Also give the coordinates of one point.
  3. Your partner sketches from your description. Compare the two drawings.
  4. Swap roles.

What to look for: the two graphs should have the same rises, falls and bends in the same places, and should agree at the one given point, but the heights of the other turning points will usually differ. That difference is exactly why derivative information gives infinitely many possible graphs.

Your turn

1

f′(x) > 0 for x < 2 and f′(x) < 0 for x > 2, and f(2) = 5. What kind of point is (2, 5)?

2

f″(x) < 0 for x < 0 and f″(x) > 0 for x > 0. What happens at x = 0?

3

Describe the shape of f on an interval where f′ < 0 and f″ > 0.

4

f′(x) = 2x − 6 and f(0) = 1. Find f(x).

5

f′(x) = 3x² and f(1) = 4. Find f(x) and describe its critical point.

6

Explain why the conditions "f′(x) > 0 for all x and f″(x) > 0 for all x" describe infinitely many functions. Give two.

More practice: warm-up

1

If f′ changes from − to + at x = 4, what is at x = 4?

2

If f(x) fits some derivative conditions, name another function that fits the same derivative conditions.

3

Shape of f where f′ > 0 and f″ < 0?

4

Shape of f where f′ < 0 and f″ < 0?

5

Is one point plus the sign of f′ enough to draw one exact graph?

More practice: on your own

1

Sketch (in words) a function with f(0) = 0, f′(x) > 0 for all x ≠ 0, f′(0) = 0, f″(x) < 0 for x < 0 and f″(x) > 0 for x > 0.

2

f′(x) = x² − 4 and f(0) = 0. Find f, its extrema and its inflection point.

3

The graph of f′ is a parabola opening up with zeros at −1 and 5. Give the x-values of the extrema and of the inflection point of f.

4

A polynomial has f′(x) = (x − 2)²(x + 1). Describe its critical points.

5

f is a cubic with f′(1) = f′(5) = 0 and f(1) = 10, f(5) = −22. Find f.

Challenge

1

Is there a function with f(x) > 0, f′(x) < 0 and f″(x) < 0 for all x? Explain.

2

Sketch, in words, the graph of f given: f(−1) = 0, f(1) = 4, f′(−1) = f′(1) = 0, f′ < 0 for x < −1, f′ > 0 for −1 < x < 1, f′ < 0 for x > 1, f″ > 0 for x < 0, f″ < 0 for x > 0. Then find a cubic that fits.

3

f′(x) = x³ − x. Without finding f, give the x-values of all extrema and inflection points of f.

Check your understanding

Quiz

1.

f′(x) = 2x and f(1) = 5. What is f(x)?

2.

Why do the conditions "f′ > 0 and f″ < 0 on (0, ∞)" describe infinitely many graphs?

3.

The graph of f′ touches the x-axis at x = 4 without crossing. At x = 4, f has:

4.

The graph of f′ has a local maximum at x = −2. At x = −2, the graph of f has:

5.

On an interval, f′ < 0 and f″ > 0. Which phrase fits f?

6.

f′(x) = (x − 1)(x − 3) and f(0) = 0. What is the local minimum point?

For parents and teachers

Big idea (B1.4): knowing where f′ and f″ are positive or negative tells you the shape of a graph on every interval, but not its exact height or steepness. So many graphs fit, and one known point plus the exact formula for f′ is needed to pin one down.

How to help: play the "Describe it, draw it" game. Your teen should be able to explain why your two drawings can differ and still both be correct.

Common difficulties:

  • Treating the graph of f′ as if it were f (a high point on f′ is an inflection point of f, not a maximum).
  • Forgetting the + C when working backwards from f′ to f, or forgetting to use the given point to find C.
  • Drawing a sharp corner at an extremum of a smooth function.