Unit 5 · Topic 5.8 Beta

Sketching f, f' and f''

Sketch a derivative feature by feature: where f rises, f′ is positive; where f is flat at a peak or valley, f′ crosses zero; where f is concave up, f′ rises; where f inflects, f′ turns; where f has a corner, f′ jumps.

Practice 2: Connecting Representations

Question set for this topic

Part 1 · Hook

Why this matters

A pilot's instruments show altitude on one dial and climb rate on another. An experienced pilot can glance at one and picture the other: when the altimeter peaks, the climb rate reads zero. Sketching f′f' from ff, and ff from f′f', is the same skill, and the exam tests it with graphs every year.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If ff is concave up on an interval, then f′f' is

  1. increasing there
  2. positive there
  3. decreasing there
  4. zero there
Show the answer

Concave up: f″>0f'' > 0, so f′f' increases.

  • Correct: increasing there:
  • positive there:
  • decreasing there:
  • zero there:

2. At a relative maximum of a differentiable ff, f′f'

  1. changes from positive to negative
  2. changes from negative to positive
  3. is positive
  4. is undefined
Show the answer

The First Derivative Test.

  • Correct: changes from positive to negative:
  • changes from negative to positive:
  • is positive:
  • is undefined:

3. f′(x)>0f'(x) > 0 on (a,b)(a, b) means ff is

  1. increasing on (a,b)(a, b)
  2. concave up on (a,b)(a, b)
  3. positive on (a,b)(a, b)
  4. decreasing on (a,b)(a, b)
Show the answer

The sign of f′f' gives direction.

  • Correct: increasing on (a,b)(a, b):
  • concave up on (a,b)(a, b):
  • positive on (a,b)(a, b):
  • decreasing on (a,b)(a, b):

Part 4 · See it

See it first

Top: f(x) = x cubed over 3 minus x squared minus 3x plus 2, with a peak at x = -1, a valley at x = 3 and an inflection point at x = 1. Bottom: f'(x) = x squared minus 2x minus 3. f' is zero at x = -1 and x = 3, where f turns, and lowest at x = 1, where f changes concavity; f' is positive where f rises and negative where f falls.
Line the two graphs up: where ff peaks or bottoms out, f′f' crosses zero; where ff inflects, f′f' turns; where ff rises, f′f' is above the axis. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. ff is rising (falling) on an intervalthe graph of f′f' is above (below) the xx-axis there
  2. ff has a peak or valley with a flat tangent linethe graph of f′f' crosses the xx-axis there
  3. ff is concave up (down)the graph of f′f' is rising (falling)
  4. ff has a point of inflectionthe graph of f′f' has a peak or valley there
  5. ff has a cornerthe graph of f′f' jumps, with no value at that point

Part 6 · Key ideas

Key ideas

  • Read a graph of ff for features, not values: where it rises, falls, peaks, bends. Each feature becomes a feature of the graph of the derivative.
  • ff (sign of slope) →\to f′f' (sign of value). ff (concavity) →\to f′f' (rising or falling). Inflection of ff →\to peak or valley of f′f'.
  • Going the other way, from f′f' to ff: zeros of f′f' where its sign changes are extrema of ff; turns of f′f' are inflection points of ff; a constant f′f' means ff is a straight line.
  • The same pattern links f′f' and f″f'': the slope of the graph of f′f' is f″f''.

Part 7 · Misconception

A common mistake

The wrong idea: The graph of f′f' looks like the graph of ff, shifted or flattened.

What actually happens: f′f' records slopes, not heights. A peak of ff is a zero of f′f', and a peak of f′f' is the steepest point of ff, an inflection point. Build the sketch feature by feature: signs first, then where f′f' rises or falls.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of f(x)=x44−2x2f(x) = \frac{x^4}{4} - 2x^2

−3−2−1123−5−4−3−2−11234xyf

1. Where does the graph of f′f' cross or touch the xx-axis?

  1. At x=−2x = -2, x=0x = 0 and x=2x = 2
  2. At x=−22x = -2\sqrt{2}, x=0x = 0 and x=22x = 2\sqrt{2}
  3. At x≈−1.15x \approx -1.15 and x≈1.15x \approx 1.15
  4. At x=−3x = -3 and x=3x = 3
Show the answer

f′(x)=0f'(x) = 0 where the graph of ff has a flat tangent line: at its turning points x=−2x = -2, 00, 22. Check: f′(x)f'(x)=x3−4x= x^3 - 4x=x(x−2)(x+2)= x(x - 2)(x + 2).

  • Correct: At x=−2x = -2, x=0x = 0 and x=2x = 2: The peaks and valleys of ff, where its tangent lines are flat.
  • At x=−22x = -2\sqrt{2}, x=0x = 0 and x=22x = 2\sqrt{2}: Those are the zeros of ff, where the graph of ff crosses the axis.
  • At x≈−1.15x \approx -1.15 and x≈1.15x \approx 1.15: Those are the inflection points of ff, where f′f' turns.
  • At x=−3x = -3 and x=3x = 3: The ends of the graph are not where ff is flat.

2. On which intervals is the graph of f′f' above the xx-axis?

  1. (−3,−2)(-3, -2) and (0,2)(0, 2)
  2. (−1.15,1.15)(-1.15, 1.15)
  3. (−3,−2)(-3, -2) and (2,3)(2, 3)
  4. (−2,0)(-2, 0) and (2,3)(2, 3)
Show the answer

f′>0f' > 0 exactly where ff is increasing: on (−2,0)(-2, 0) and (2,3)(2, 3).

  • (−3,−2)(-3, -2) and (0,2)(0, 2): Here ff falls, so f′<0f' < 0.
  • (−1.15,1.15)(-1.15, 1.15): There ff is concave down, which is about the slope of f′f', not its sign.
  • (−3,−2)(-3, -2) and (2,3)(2, 3): ff falls on (−3,−2)(-3, -2), so f′f' is negative there.
  • Correct: (−2,0)(-2, 0) and (2,3)(2, 3): Where ff rises, its derivative is positive.

3. Where does the graph of f′f' have its relative maximum and minimum?

  1. At the relative extrema of ff, x=−2x = -2, 00, 22
  2. At the endpoints x=−3x = -3 and x=3x = 3
  3. At the inflection points of ff, x=±23≈±1.15x = \pm\frac{2}{\sqrt{3}} \approx \pm 1.15
  4. At the zeros of ff, x=0x = 0 and x=±22x = \pm 2\sqrt{2}
Show the answer

f″(x)=3x2−4f''(x) = 3x^2 - 4 changes sign at x=±23x = \pm\frac{2}{\sqrt{3}}, where ff changes concavity. There f′f' changes from increasing to decreasing or back, so the graph of f′f' peaks at x≈−1.15x \approx -1.15 and bottoms out at x≈1.15x \approx 1.15.

  • At the relative extrema of ff, x=−2x = -2, 00, 22: There f′=0f' = 0; those are zeros of the graph of the derivative, not its turns.
  • At the endpoints x=−3x = -3 and x=3x = 3: An endpoint is not a relative extremum.
  • Correct: At the inflection points of ff, x=±23≈±1.15x = \pm\frac{2}{\sqrt{3}} \approx \pm 1.15: f′f' turns where its own slope f″f'' changes sign, which is where ff changes concavity.
  • At the zeros of ff, x=0x = 0 and x=±22x = \pm 2\sqrt{2}: The zeros of ff say nothing about f′f'.

4. At x=1x = 1, what does the graph of f′f' look like?

  1. Below the xx-axis and rising
  2. Below the xx-axis and falling
  3. Above the xx-axis and falling
  4. Above the xx-axis and rising
Show the answer

At x=1x = 1 the graph of ff is going down and bending down. So f′(1)<0f'(1) < 0 (f′(1)=1−4=−3f'(1) = 1 - 4 = -3) and f′f' is decreasing (f″(1)=−1<0f''(1) = -1 < 0).

  • Below the xx-axis and rising: ff is concave down at x=1x = 1, so f′f' is falling.
  • Correct: Below the xx-axis and falling: ff is falling (f′<0f' < 0) and concave down (f′f' decreasing) at x=1x = 1.
  • Above the xx-axis and falling: ff is decreasing at x=1x = 1, so f′f' is negative.
  • Above the xx-axis and rising: Both features are reversed.

5. The graph of f′f' is a line with positive slope that crosses the xx-axis at x=1x = 1. Which describes the graph of ff?

  1. A parabola opening upward with its vertex at x=1x = 1
  2. A parabola opening downward with its vertex at x=1x = 1
  3. A line with positive slope through (1,0)(1, 0)
  4. A cubic with a point of inflection at x=1x = 1
Show the answer

f′(x)=m(x−1)f'(x) = m(x - 1) with m>0m > 0, so f(x)=m2(x−1)2+kf(x) = \frac{m}{2}(x - 1)^2 + k for some constant kk: decreasing then increasing, concave up, vertex at x=1x = 1.

  • Correct: A parabola opening upward with its vertex at x=1x = 1: f′<0f' < 0 before 1 and f′>0f' > 0 after (a minimum at 1), and f′f' increasing means concave up.
  • A parabola opening downward with its vertex at x=1x = 1: That would need f′f' to go from positive to negative.
  • A line with positive slope through (1,0)(1, 0): A line is f′f' here; ff is one degree higher.
  • A cubic with a point of inflection at x=1x = 1: f″f'' is a positive constant, so the concavity never changes.

6. The graph of ff is concave down on (a,b)(a, b). What must be true of the graph of f′f' there?

  1. It is below the xx-axis
  2. It is concave down
  3. It is increasing
  4. It is decreasing
Show the answer

Concave down on (a,b)(a, b) means the slopes of ff decrease there: the graph of the derivative is decreasing.

  • It is below the xx-axis: The sign of f′f' gives the direction of ff, not its concavity.
  • It is concave down: The concavity of f′f' is about f‴f'''.
  • It is increasing: That would be concave up.
  • Correct: It is decreasing: Concave down means f″<0f'' < 0, so f′f' is decreasing.

Part 9 · Summary

Summary

Sketch a derivative feature by feature: where ff rises, f′f' is positive; where ff is flat at a peak or valley, f′f' crosses zero; where ff is concave up, f′f' rises; where ff inflects, f′f' turns; where ff has a corner, f′f' jumps. Read the same links backward to sketch ff from f′f'.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections