Unit 5 · Topic 5.4 Beta

The First Derivative Test

At a critical point, look at the sign of f′ on each side.

Practice 3: Justification

Question set for this topic

Part 1 · Hook

Why this matters

A cyclist on a hilly road knows she has reached the top of a climb when the road stops going up and starts going down. A flat stretch alone does not prove it: the road may level off and keep climbing. The First Derivative Test turns that common sense into a rule you can state and justify.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

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

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

A positive derivative means rising tangent lines.

  • Correct: increasing:
  • decreasing:
  • concave up:
  • constant:

2. Which points are critical points of ff?

  1. Interior points where f′=0f' = 0 or f′f' does not exist
  2. Points where f=0f = 0
  3. The endpoints only
  4. Points where ff is largest
Show the answer

Relative extrema can occur only at critical points.

  • Correct: Interior points where f′=0f' = 0 or f′f' does not exist:
  • Points where f=0f = 0:
  • The endpoints only:
  • Points where ff is largest:

Part 4 · See it

See it first

Three graphs, each with a critical point at x = c. Left: f' changes from positive to negative, the graph rises then falls, a relative maximum. Middle: f' changes from negative to positive, a relative minimum. Right: f' is positive on both sides, the graph rises through a flat point, no relative extremum.
At a critical point cc, the change in the sign of f′f' decides: ++ to −- is a peak, −- to ++ a valley, and no change means no extremum. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. f′f' changes from positive to negative at a critical point cc (and ff is continuous at cc)ff rises up to cc and falls after it, so f(c)f(c) is a relative maximum
  2. f′f' changes from negative to positive at ccff falls then rises, so f(c)f(c) is a relative minimum
  3. f′f' has the same sign on both sides of ccff keeps rising (or falling) through cc: no relative extremum, even if f′(c)=0f'(c) = 0
  4. You state the resultwrite the sentence: "f′f' changes from negative to positive at x=cx = c, so ff has a relative minimum there"

Part 6 · Key ideas

Key ideas

  • First Derivative Test: at a critical point cc where ff is continuous, if f′f' changes from ++ to −-, f(c)f(c) is a relative maximum; from −- to ++, a relative minimum; no change, neither.
  • It works where f′f' does not exist (corners, cusps) as well as where f′=0f' = 0.
  • A sign chart of f′f' organizes the work, but it is not a justification by itself: write the sentence.
  • On a graph of f′f': a relative maximum of ff is where the graph of f′f' crosses the axis going down; a minimum, crossing going up. Touching without crossing gives nothing.

Part 7 · Misconception

A common mistake

The wrong idea: f′(c)=0f'(c) = 0, so ff has a relative maximum or minimum at cc.

What actually happens: The scoring guidelines give no credit for "f′(c)=0f'(c) = 0" alone. The point needs a sign change, stated with its direction: "f′f' changes from positive to negative at x=cx = c." For x3x^3 at 0, f′=0f' = 0 but there is no extremum.

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)(x−2)220= \frac{x(x + 3)(x - 2)^2}{20} for −3.8≤x≤3.2-3.8 \le x \le 3.2. Its zeros are x=−3x = -3, x=0x = 0 and x=2x = 2, where the graph touches the axis without crossing.
−4−3−2−1123−2−112345xyf'

1. At what value of xx does ff have a relative maximum?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

f′f' changes from positive to negative at x=−3x = -3, so ff has a relative maximum there.

  • Answer: -3

2. What happens to ff at x=2x = 2?

  1. A relative minimum, because f′(2)=0f'(2) = 0
  2. A relative maximum, because f′f' decreases down to 0 just before x=2x = 2
  3. A relative minimum, because f′f' has a low point at x=2x = 2
  4. No relative extremum, because f′f' does not change sign at x=2x = 2
Show the answer

f′(x)>0f'(x) > 0 on both sides of x=2x = 2 (the graph touches the axis there), so ff is increasing through x=2x = 2 and has no relative extremum.

  • A relative minimum, because f′(2)=0f'(2) = 0: f′(2)=0f'(2) = 0 makes 2 a critical point, not an extremum.
  • A relative maximum, because f′f' decreases down to 0 just before x=2x = 2: f′f' is positive on both sides; ff rises through x=2x = 2.
  • A relative minimum, because f′f' has a low point at x=2x = 2: A low point of f′f' says something about concavity, and f′f' never goes negative here.
  • Correct: No relative extremum, because f′f' does not change sign at x=2x = 2: f′>0f' > 0 just left and just right of 2: ff keeps rising.

3. Which justification earns credit for "ff has a relative minimum at x=0x = 0"?

  1. "f′(0)=0f'(0) = 0."
  2. "f′(x)<0f'(x) < 0 for −3<x<0-3 < x < 0."
  3. "f′f' changes from negative to positive at x=0x = 0."
  4. "The graph crosses the xx-axis at x=0x = 0."
Show the answer

First Derivative Test: f′f' changes from negative to positive at x=0x = 0, so ff has a relative minimum there.

  • "f′(0)=0f'(0) = 0.": A zero of f′f' alone does not decide: see x=2x = 2.
  • "f′(x)<0f'(x) < 0 for −3<x<0-3 < x < 0.": Half the story: f′f' must also be positive just after 0.
  • Correct: "f′f' changes from negative to positive at x=0x = 0.": It names f′f' and the direction of the sign change.
  • "The graph crosses the xx-axis at x=0x = 0.": Which graph, and crossing in which direction? Name f′f' and the signs.

4. At which values of xx does ff have a relative minimum? Select each one.

  1. x=−3x = -3
  2. x=0x = 0
  3. x=2x = 2
  4. x≈−1.7x \approx -1.7, where f′f' is lowest
Show the answer

Only at x=0x = 0 does f′f' change from negative to positive.

  • x=−3x = -3: f′f' goes from positive to negative there: a relative maximum.
  • Correct: x=0x = 0: f′f' goes from negative to positive.
  • x=2x = 2: f′f' does not change sign.
  • x≈−1.7x \approx -1.7, where f′f' is lowest: f′f' is negative on both sides of −1.7-1.7; ff is still falling.

5. Find the relative maximum value of f(x)=x3−3x2−9x+4f(x) = x^3 - 3x^2 - 9x + 4.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

f′(x)f'(x)=3x2−6x−9= 3x^2 - 6x - 9=3(x−3)(x+1)= 3(x - 3)(x + 1). f′f' changes from positive to negative at x=−1x = -1, so the relative maximum value is f(−1)f(-1)=−1−3+9+4= -1 - 3 + 9 + 4=9= 9.

  • Answer: 9

6. Which function has f′(0)=0f'(0) = 0 but no relative extremum at x=0x = 0?

  1. f(x)=x2f(x) = x^2
  2. f(x)=x4−1f(x) = x^4 - 1
  3. f(x)=x3f(x) = x^3
  4. f(x)=3−x2f(x) = 3 - x^2
Show the answer

For x3x^3, f′(x)=3x2≥0f'(x) = 3x^2 \ge 0 on both sides of 0, so ff rises through 0. A zero of f′f' without a sign change is not an extremum.

  • f(x)=x2f(x) = x^2: f′(x)=2xf'(x) = 2x changes from negative to positive: a minimum.
  • f(x)=x4−1f(x) = x^4 - 1: f′(x)=4x3f'(x) = 4x^3 changes from negative to positive: a minimum.
  • Correct: f(x)=x3f(x) = x^3: f′(x)=3x2f'(x) = 3x^2 is positive on both sides of 0: no sign change.
  • f(x)=3−x2f(x) = 3 - x^2: f′(x)=−2xf'(x) = -2x changes from positive to negative: a maximum.

Part 9 · Summary

Summary

At a critical point, look at the sign of f′f' on each side. ++ to −-: relative maximum. −- to ++: relative minimum. No change: no extremum. The test works where f′f' is undefined too. Justify in words that name f′f' and the direction of its sign change; a sign chart alone does not earn the point.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections