Unit 5 · Topic 5.6 Beta

Concavity and Points of Inflection

The sign of f″ gives the concavity: f″ > 0 (equivalently f′ increasing) means concave up, f″ < 0 means concave down.

Practice 3: Justification

Question set for this topic

Part 1 · Hook

Why this matters

News reports say "cases are still rising, but more slowly." That sentence is about concavity: the count is increasing, but its rate of increase is falling. The moment the rate stops speeding up and starts slowing down is a point of inflection, and it is often the most watched point on a graph.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A graph that is concave up lies

  1. above its tangent lines
  2. below its tangent lines
  3. on its tangent lines
  4. above the xx-axis
Show the answer

Concave up: the curve bends upward away from each tangent line.

  • Correct: above its tangent lines:
  • below its tangent lines:
  • on its tangent lines:
  • above the xx-axis:

2. If f(x)=x3f(x) = x^3, then f″(x)=f''(x) =

  1. 6x6x
  2. 3x23x^2
  3. 66
  4. xx
Show the answer

f′(x)=3x2f'(x) = 3x^2, and f″(x)=6xf''(x) = 6x.

  • Correct: 6x6x:
  • 3x23x^2:
  • 66:
  • xx:

3. If f′(x)>0f'(x) > 0 on an interval, then ff is

  1. increasing there
  2. concave up there
  3. decreasing there
  4. concave down there
Show the answer

The sign of f′f' gives direction.

  • Correct: increasing there:
  • concave up there:
  • decreasing there:
  • concave down there:

Part 4 · See it

See it first

Top: f(x) = x cubed minus 3x squared plus 2 with short tangent lines; for x < 1 the graph is concave down and its tangent lines lie above it, for x > 1 it is concave up and its tangent lines lie below it; the point of inflection is (1, 0). Bottom: f'(x) = 3x squared minus 6x, decreasing for x < 1 and increasing for x > 1, lowest at x = 1, where f''(x) = 6x - 6 changes sign.
Concave down: tangent lines above the curve and f′f' decreasing. Concave up: tangent lines below and f′f' increasing. The change happens at the point of inflection, where f′f' is lowest. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. f″(x)>0f''(x) > 0 on an intervalf′f' is increasing there, the slopes grow, and the graph of ff is concave up
  2. f″(x)<0f''(x) < 0 on an intervalf′f' is decreasing, the slopes shrink, and the graph is concave down
  3. f″f'' changes sign at x=cx = c (and ff is continuous there)the concavity changes, so (c,f(c))(c, f(c)) is a point of inflection
  4. f″(c)=0f''(c) = 0 but f″f'' keeps its sign (as for x4x^4)no change of concavity, so no point of inflection

Part 6 · Key ideas

Key ideas

  • Concave up on an interval: f′f' is increasing there (f″>0f'' > 0). Concave down: f′f' is decreasing (f″<0f'' < 0).
  • Point of inflection: a point on the graph where the concavity changes, that is, where f″f'' changes sign, or where f′f' changes from increasing to decreasing or back.
  • On a graph of f′f': inflection points of ff are at the high and low points of f′f', not at its zeros.
  • Candidates are where f″=0f'' = 0 or f″f'' does not exist; then check the sign of f″f'' on each side.

Part 7 · Misconception

A common mistake

The wrong idea: f″(c)=0f''(c) = 0, so the graph of ff has a point of inflection at x=cx = c.

What actually happens: f″(c)=0f''(c) = 0 only makes cc a candidate. The concavity must actually change: f″f'' must change sign. f(x)=x4f(x) = x^4 has f″(0)=0f''(0) = 0 but is concave up on both sides. On a graph of f′f', also do not confuse zeros of f′f' (extrema of ff) with turns of f′f' (inflection points).

Part 8 · Check yourself

Check yourself

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

Graph

The graph of f′f', the derivative of ff

The graph shows f′(x)f'(x)=x(x−3)22−1= \frac{x(x - 3)^2}{2} - 1 for −0.5≤x≤4.5-0.5 \le x \le 4.5. The graph of f′f' has a high point at x=1x = 1 and a low point at x=3x = 3, and crosses the xx-axis at x=2−3≈0.27x = 2 - \sqrt{3} \approx 0.27, x=2x = 2 and x=2+3≈3.73x = 2 + \sqrt{3} \approx 3.73.
−112345−5−4−3−2−112345xyf'

1. On which interval is the graph of ff concave down?

  1. (1,3)(1, 3)
  2. (2,3.73)(2, 3.73), where f′<0f' < 0
  3. (−0.5,1)(-0.5, 1) and (3,4.5)(3, 4.5)
  4. (0.27,2)(0.27, 2), where f′>0f' > 0
Show the answer

ff is concave down where f′f' is decreasing: between the high point of f′f' at x=1x = 1 and its low point at x=3x = 3.

  • Correct: (1,3)(1, 3): f′f' is decreasing on (1,3)(1, 3), so f″<0f'' < 0 there.
  • (2,3.73)(2, 3.73), where f′<0f' < 0: f′<0f' < 0 means ff is decreasing, not concave down.
  • (−0.5,1)(-0.5, 1) and (3,4.5)(3, 4.5): There f′f' is increasing: ff is concave up.
  • (0.27,2)(0.27, 2), where f′>0f' > 0: f′>0f' > 0 means ff is increasing. Concavity comes from whether f′f' rises or falls.

2. At which values of xx does the graph of ff have a point of inflection?

  1. x=0.27x = 0.27, x=2x = 2 and x=3.73x = 3.73
  2. x=2x = 2, where the graph of f′f' crosses between its turns
  3. x=−0.5x = -0.5 and x=4.5x = 4.5, the endpoints
  4. x=1x = 1 and x=3x = 3
Show the answer

Points of inflection are where f″f'' changes sign, that is, where f′f' changes from increasing to decreasing or back: at the relative extrema of f′f', x=1x = 1 and x=3x = 3.

  • x=0.27x = 0.27, x=2x = 2 and x=3.73x = 3.73: Those are zeros of f′f': candidates for extrema of ff, not inflection points.
  • x=2x = 2, where the graph of f′f' crosses between its turns: x=2x = 2 is a zero of f′f'; f′f' is decreasing on both sides of it.
  • x=−0.5x = -0.5 and x=4.5x = 4.5, the endpoints: An endpoint cannot be a point of inflection: concavity must change on both sides.
  • Correct: x=1x = 1 and x=3x = 3: f′f' changes from increasing to decreasing at 1 and from decreasing to increasing at 3.

3. Which describes the graph of ff at x=2.5x = 2.5?

  1. Decreasing and concave up
  2. Increasing and concave down
  3. Decreasing and concave down
  4. Increasing and concave up
Show the answer

f′(2.5)≈−0.69<0f'(2.5) \approx -0.69 < 0, so ff is decreasing; f′f' is decreasing on (1,3)(1, 3), so f″<0f'' < 0 and ff is concave down.

  • Decreasing and concave up: f′f' is still decreasing at 2.5; it turns up only at x=3x = 3.
  • Increasing and concave down: f′(2.5)f'(2.5)=2.5(0.25)2−1<0= \frac{2.5(0.25)}{2} - 1 < 0, so ff is decreasing.
  • Correct: Decreasing and concave down: f′(2.5)<0f'(2.5) < 0 (below the axis) and f′f' is falling there.
  • Increasing and concave up: Both conditions are reversed.

4. Which justification earns credit for "the graph of ff has a point of inflection at x=3x = 3"?

  1. "f′f' changes sign at x=3x = 3."
  2. "f′f' changes from decreasing to increasing at x=3x = 3."
  3. "f″(3)=0f''(3) = 0."
  4. "ff changes from decreasing to increasing at x=3x = 3."
Show the answer

An inflection point needs a change of concavity: f″f'' changes sign, or equivalently f′f' changes from decreasing to increasing (or back). At x=3x = 3, f′f' has a low point.

  • "f′f' changes sign at x=3x = 3.": f′(3)=−1f'(3) = -1; f′f' does not change sign there. A sign change of f′f' would be about extrema.
  • Correct: "f′f' changes from decreasing to increasing at x=3x = 3.": That means f″f'' changes from negative to positive: concavity changes.
  • "f″(3)=0f''(3) = 0.": f″(3)=0f''(3) = 0 alone does not prove a change of concavity.
  • "ff changes from decreasing to increasing at x=3x = 3.": f′<0f' < 0 on both sides of 3, so ff is decreasing on both sides; and that would be about extrema anyway.

5. For f(x)=x4f(x) = x^4, f″(0)=0f''(0) = 0. Is there a point of inflection at x=0x = 0?

  1. Yes, because f″(0)=0f''(0) = 0
  2. Yes, because the graph is flat at x=0x = 0
  3. No, because f′(0)=0f'(0) = 0
  4. No, because f″(x)=12x2≥0f''(x) = 12x^2 \ge 0 on both sides, so the concavity does not change
Show the answer

f″(x)=12x2f''(x) = 12x^2 is positive on both sides of 0. The concavity does not change, so there is no point of inflection, even though f″(0)=0f''(0) = 0.

  • Yes, because f″(0)=0f''(0) = 0: f″(c)=0f''(c) = 0 makes a candidate only; f″f'' must change sign.
  • Yes, because the graph is flat at x=0x = 0: Flatness is about f′f', not concavity.
  • No, because f′(0)=0f'(0) = 0: The right answer for a wrong reason: f′(0)=0f'(0) = 0 is about extrema.
  • Correct: No, because f″(x)=12x2≥0f''(x) = 12x^2 \ge 0 on both sides, so the concavity does not change: The graph is concave up on both sides of 0.

6. N(t)N(t) is the number of people who have heard a rumor tt days after it starts. N′(t)N'(t) increases for 0<t<30 < t < 3 and decreases for t>3t > 3. What happens at t=3t = 3?

  1. The most people have heard the rumor
  2. The rumor stops spreading
  3. The rumor is spreading fastest; the graph of NN has a point of inflection
  4. The rumor is spreading slowest; NN has a relative minimum
Show the answer

N′N' changes from increasing to decreasing at t=3t = 3, so N″N'' changes from positive to negative: NN has a point of inflection, and the rate N′N' is greatest there.

  • The most people have heard the rumor: NN itself can keep growing after t=3t = 3; only its rate slows.
  • The rumor stops spreading: N′(3)N'(3) is its largest value, not 0.
  • Correct: The rumor is spreading fastest; the graph of NN has a point of inflection: N′N', the spreading rate, peaks at t=3t = 3, where NN changes from concave up to concave down.
  • The rumor is spreading slowest; NN has a relative minimum: N′N' is largest at 3, not smallest.

Part 9 · Summary

Summary

The sign of f″f'' gives the concavity: f″>0f'' > 0 (equivalently f′f' increasing) means concave up, f″<0f'' < 0 means concave down. A point of inflection is where the concavity changes, so f″f'' must change sign; f″(c)=0f''(c) = 0 alone is not enough. On a graph of f′f', look at where f′f' turns.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections