Unit 5 · Topic 5.3 Beta

Where a Function Increases or Decreases

The sign of the derivative gives the direction of a function: f′ > 0 on an interval means f is increasing there, f′ < 0 means decreasing.

Practice 3: Justification

Question set for this topic

Part 1 · Hook

Why this matters

A doctor reads a heart monitor by its slope: a line climbing means the signal is rising, a line falling means it is dropping. Calculus reads every graph that way. The derivative is the slope, so its sign tells you where the function rises and where it falls, even when you cannot see the function itself.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If f′(2)=−3f'(2) = -3, the tangent line to ff at x=2x = 2

  1. slopes downward
  2. slopes upward
  3. is horizontal
  4. is vertical
Show the answer

A negative slope falls from left to right.

  • Correct: slopes downward:
  • slopes upward:
  • is horizontal:
  • is vertical:

2. Where is f′(x)=3x2−12f'(x) = 3x^2 - 12 equal to 0?

  1. x=±2x = \pm 2
  2. x=4x = 4
  3. x=±4x = \pm 4
  4. x=0x = 0
Show the answer

3x2=123x^2 = 12, so x2=4x^2 = 4.

  • Correct: x=±2x = \pm 2:
  • x=4x = 4:
  • x=±4x = \pm 4:
  • x=0x = 0:

3. The critical points of ff are interior points where

  1. f′(x)=0f'(x) = 0 or f′(x)f'(x) does not exist
  2. f(x)=0f(x) = 0
  3. f″(x)=0f''(x) = 0
  4. ff is not continuous
Show the answer

They split the domain into pieces where f′f' keeps one sign.

  • Correct: f′(x)=0f'(x) = 0 or f′(x)f'(x) does not exist:
  • f(x)=0f(x) = 0:
  • f″(x)=0f''(x) = 0:
  • ff is not continuous:

Part 4 · See it

See it first

Top: f(x) = x cubed minus 3x. Bottom, on the same x-scale: f'(x) = 3x squared minus 3. Where f' is above the x-axis (x < -1 and x > 1) f is increasing; where f' is below the axis (-1 < x < 1) f is decreasing. Dashed lines at x = -1 and 1 join the zeros of f' to the turning points of f.
Read the sign of f′f' straight up to ff: above the axis, ff rises; below it, ff falls. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. f′(x)>0f'(x) > 0 at each point of an intervaleach tangent line slopes up, so ff is increasing on that interval
  2. f′(x)<0f'(x) < 0 at each point of an intervalff is decreasing on that interval
  3. f′f' can change sign only at a zero or where f′f' does not exist (if f′f' has no jumps)the critical points split the line into intervals, and one test value per interval gives the sign
  4. You are given the graph of f′f'read the sign (above or below the axis), not whether the graph of f′f' goes up or down

Part 6 · Key ideas

Key ideas

  • Increasing on an interval: f(x1)<f(x2)f(x_1) < f(x_2) whenever x1<x2x_1 < x_2 in the interval. If f′(x)>0f'(x) > 0 on the interval, ff is increasing there; if f′(x)<0f'(x) < 0, decreasing.
  • Method: find f′f', find where it is 0 or undefined, test the sign of f′f' on each interval between those points.
  • On a graph of f′f': above the xx-axis means ff increasing; below means ff decreasing.
  • Justify with the sign of f′f' by name: "f′(x)>0f'(x) > 0 for −2<x<1-2 < x < 1, so ff is increasing there."

Part 7 · Misconception

A common mistake

The wrong idea: On a graph of f′f', ff is increasing wherever the graph of f′f' is going up.

What actually happens: The direction of ff comes from the sign of f′f': above the axis, ff increases, even where the graph of f′f' is falling. Whether f′f' itself rises or falls tells you about the concavity of ff, a later topic.

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+2)(x−1)(x−4)5= \frac{(x + 2)(x - 1)(x - 4)}{5} for −3≤x≤5-3 \le x \le 5. The function ff is differentiable on this interval.
−3−2−112345−6−4−2246xyf'

1. On which intervals is ff increasing?

  1. (−2,1)(-2, 1) and (4,5)(4, 5)
  2. (−3,−2)(-3, -2) and (1,4)(1, 4)
  3. (−3,−0.73)(-3, -0.73) and (2.73,5)(2.73, 5), where f′f' rises
  4. (−2,4)(-2, 4), between the outer zeros of f′f'
Show the answer

ff is increasing where f′(x)>0f'(x) > 0: on (−2,1)(-2, 1) and on (4,5)(4, 5), where the graph of f′f' lies above the xx-axis.

  • Correct: (−2,1)(-2, 1) and (4,5)(4, 5): f′>0f' > 0 exactly there: the graph of f′f' is above the xx-axis.
  • (−3,−2)(-3, -2) and (1,4)(1, 4): Here f′<0f' < 0: these are where ff is decreasing.
  • (−3,−0.73)(-3, -0.73) and (2.73,5)(2.73, 5), where f′f' rises: That is where f′f' is increasing. The sign of f′f', not its direction, tells whether ff increases.
  • (−2,4)(-2, 4), between the outer zeros of f′f': f′<0f' < 0 on (1,4)(1, 4), so ff decreases on part of this interval.

2. At x=3x = 3, is ff increasing or decreasing, and why?

  1. Increasing, because f′f' is increasing at x=3x = 3
  2. Increasing, because f′(3)≠0f'(3) \ne 0
  3. Decreasing, because f′(3)f'(3) is less than f′(1)f'(1)
  4. Decreasing, because f′(3)<0f'(3) < 0
Show the answer

f′(3)=−2<0f'(3) = -2 < 0, so ff is decreasing at x=3x = 3. The graph of f′f' is below the xx-axis there.

  • Increasing, because f′f' is increasing at x=3x = 3: f′f' is rising there, but it is still negative. The sign of f′f' decides.
  • Increasing, because f′(3)≠0f'(3) \ne 0: Nonzero does not tell the direction; negative means decreasing.
  • Decreasing, because f′(3)f'(3) is less than f′(1)f'(1): Comparing values of f′f' is about whether f′f' rises or falls, not about the sign of f′(3)f'(3).
  • Correct: Decreasing, because f′(3)<0f'(3) < 0: f′(3)f'(3)=(5)(2)(−1)5= \frac{(5)(2)(-1)}{5}=−2<0= -2 < 0.

3. On which of these intervals is ff increasing at each point? Select each one.

  1. (−2,0)(-2, 0)
  2. (1,3)(1, 3)
  3. (−3,−2)(-3, -2)
  4. (3,5)(3, 5)
Show the answer

Only (−2,0)(-2, 0) lies inside a stretch where f′>0f' > 0. The interval (3,5)(3, 5) crosses the zero of f′f' at x=4x = 4.

  • Correct: (−2,0)(-2, 0): f′>0f' > 0 on (−2,1)(-2, 1), which contains (−2,0)(-2, 0).
  • (1,3)(1, 3): f′<0f' < 0 on (1,4)(1, 4).
  • (−3,−2)(-3, -2): f′<0f' < 0 for x<−2x < -2.
  • (3,5)(3, 5): f′<0f' < 0 on (3,4)(3, 4), so ff decreases on part of (3,5)(3, 5).

4. Which is true about the values of ff?

  1. f(1)<f(4)f(1) < f(4), because f′(4)>f′(1)f'(4) > f'(1)
  2. f(1)>f(4)f(1) > f(4), because f′<0f' < 0 on (1,4)(1, 4)
  3. f(1)=f(4)f(1) = f(4), because f′(1)=f′(4)=0f'(1) = f'(4) = 0
  4. The graph of f′f' gives no way to compare them
Show the answer

f′(x)<0f'(x) < 0 for 1<x<41 < x < 4, so ff is decreasing on [1,4][1, 4] and f(4)<f(1)f(4) < f(1).

  • f(1)<f(4)f(1) < f(4), because f′(4)>f′(1)f'(4) > f'(1): f′(1)=f′(4)=0f'(1) = f'(4) = 0; and comparing f′f' values does not compare ff values.
  • Correct: f(1)>f(4)f(1) > f(4), because f′<0f' < 0 on (1,4)(1, 4): ff is decreasing on (1,4)(1, 4) and continuous, so it ends lower than it starts.
  • f(1)=f(4)f(1) = f(4), because f′(1)=f′(4)=0f'(1) = f'(4) = 0: Equal slopes at two points say nothing about equal heights.
  • The graph of f′f' gives no way to compare them: The sign of f′f' between them does compare them.

5. Which justification earns credit for "ff is increasing on (−2,1)(-2, 1)"?

  1. "f′(x)>0f'(x) > 0 for −2<x<1-2 < x < 1."
  2. "f′f' is increasing on (−2,1)(-2, 1)."
  3. "The graph goes up on (−2,1)(-2, 1)."
  4. "f′(−2)=0f'(-2) = 0 and f′(1)=0f'(1) = 0."
Show the answer

A justification names the function and the condition: f′(x)>0f'(x) > 0 on (−2,1)(-2, 1), so ff is increasing there.

  • Correct: "f′(x)>0f'(x) > 0 for −2<x<1-2 < x < 1.": It names f′f' and gives its sign on the interval.
  • "f′f' is increasing on (−2,1)(-2, 1).": False (f′f' rises then falls there), and the direction of f′f' is about concavity, not about ff increasing.
  • "The graph goes up on (−2,1)(-2, 1).": Which graph? This is the graph of f′f', and "goes up" names no function.
  • "f′(−2)=0f'(-2) = 0 and f′(1)=0f'(1) = 0.": Zeros at the ends do not tell the sign in between.

6. f′(x)=(x−3)2(x+1)f'(x) = (x - 3)^2 (x + 1). On which interval is ff decreasing?

  1. (−1,3)(-1, 3)
  2. (−∞,−1)(-\infty, -1)
  3. (3,∞)(3, \infty)
  4. (−∞,−1)(-\infty, -1) and (−1,3)(-1, 3)
Show the answer

The squared factor is never negative, so f′(x)<0f'(x) < 0 exactly when x+1<0x + 1 < 0: ff decreases on (−∞,−1)(-\infty, -1).

  • (−1,3)(-1, 3): For −1<x<3-1 < x < 3, both factors give f′>0f' > 0.
  • Correct: (−∞,−1)(-\infty, -1): (x−3)2≥0(x - 3)^2 \ge 0, so the sign of f′f' is the sign of x+1x + 1.
  • (3,∞)(3, \infty): For x>3x > 3, f′>0f' > 0.
  • (−∞,−1)(-\infty, -1) and (−1,3)(-1, 3): f′>0f' > 0 on (−1,3)(-1, 3).

Part 9 · Summary

Summary

The sign of the derivative gives the direction of a function: f′>0f' > 0 on an interval means ff is increasing there, f′<0f' < 0 means decreasing. Split the domain at the critical points, test each piece, and justify with the sign of f′f' by name. On a graph of f′f', read the sign, not the slope.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections