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Increasing, Decreasing Functions and Maximum and Minimum Values

Mathematics

Increasing, Decreasing Functions and Maximum and Minimum Values

Overview

Overview

  • This lesson covers lessons 2-1 and 2-2 of the unit "Behaviour of the Function and Curve Sketching".
  • After it you should be able to:
  • use the sign of the first derivative to find where a function increases or decreases;
  • find critical points and classify local maximum and minimum values with the first derivative test;
  • find the absolute maximum and minimum of a continuous function on a closed interval.

🌳 Concept tree

  1. Increasing
  2. Decreasing Functions and Maximum and Minimum Values

├── 2-1 (a) Monotonic functions and the tangent

│ ├── Definitions

│ └── Link with the tangent

  • ├── 2-1 (b) The first-derivative test for monotony

│ ├── Rule

│ ├── Steps

│ ├── Examples

│ └── Notes

├── 2-2 (a) Critical points

│ ├── Definition

│ └── Reading a graph

├── 2-2 (b) Local maximum and minimum values

│ ├── Definitions

│ └── Necessary condition

  • ├── 2-2 (c) First-derivative test for local extrema

│ ├── Rule

│ └── Steps

  • ├── 2-2 (d) Absolute maximum and minimum on a closed interval

│ ├── Definition

│ ├── Existence

│ ├── Method

│ └── Remarks

  • ├── 2-2 (e) Finding unknown constants from extrema

│ └── Idea

└── 2-2 (f) Summary

└── Comparison of the tests

Increasing, Decreasing Functions and Maximum and Minimum Values · 1 / 9
2-1 (a) Monotonic functions and the tangent

2-1 (a) Monotonic functions and the tangent

Definitions

  • A function f is increasing on an interval if its graph rises from left to right: x1<x2 gives f(x1)<f(x2).
  • It is decreasing if its graph falls x1<x2 gives f(x1)>f(x2).
  • Reading a graph gives the intervals, but differentiation gives them exactly.
  • Where the function increases, the tangent makes an acute angle with the positive x-axis, so its slope f′(x) is positive.
  • Where it decreases, the tangent makes an obtuse angle, so f′(x) is negative.
  • Where the tangent is horizontal, f′(x)=0.
Increasing, Decreasing Functions and Maximum and Minimum Values · 2 / 9
2-1 (b) The first-derivative test for monotony

2-1 (b) The first-derivative test for monotony

Rule

  • Let f be differentiable on the open interval ]a,b[.
  • If f′(x)>0 for every x∈ ]a,b[, then f is increasing on ]a,b[.
  • If f′(x)<0 for every x∈ ]a,b[, then f is decreasing on ]a,b[.
  • Find the domain of f and check where it is continuous and differentiable.
  • Find f′(x).
  • Solve f′(x)=0 and find where f′(x) does not exist; these values separate the intervals.
  • Make a table of signs of f′(x) on each interval (test one value in each).
  • State the intervals of increase and decrease.

Examples

✍️ Worked examples

  • Example 1 Find the intervals of increase and decrease of f(x)=x3−3x+2.
  • Solution:
  • f is a polynomial, so it is continuous and differentiable on R.
  • f′(x)=3x2−3=3(x−1)(x+1).
  • f′(x)=0 at x=−1 and x=1.
  • Sign of f′ positive on ]−∞,−1[, negative on ]−1,1[, positive on ]1,∞[.
  • So f is increasing on ]−∞,−1[ and on ]1,∞[, and decreasing on ]−1,1[.
  • Example 2 Find the intervals of monotony of f(x)=x+2sin⁡x for 0<x<2π.
  • Solution:
  • f′(x)=1+2cos⁡x.
  • f′(x)=0 gives cos⁡x=−12, so x=2π3 or x=4π3.
  • Test points at x=π2, f′=1>0; at x=π, f′=−1<0; at x=3π2, f′=1>0.
  • Increasing on ]0,2π3[ and ]4π3,2π[; decreasing on ]2π3,4π3[.
  • Example 3 Find the intervals of monotony of h(x)=2ln⁡x−x2.
  • Solution:
  • The domain is ]0,∞[.
  • h′(x)=2x−2x=2(1−x2)x.
  • h′(x)=0 gives x=1 (since x>0).
  • For 0<x<1, h′>0; for x>1, h′<0.
  • h is increasing on ]0,1[ and decreasing on ]1,∞[.
  • Example 4 Find the intervals of monotony of f(x)=xx2+1.
  • Solution:
  • f′(x)=(x2+1)−2x2(x2+1)2=1−x2(x2+1)2.
  • The denominator is always positive, so the sign depends on 1−x2.
  • f′(x)>0 on ]−1,1[ and f′(x)<0 on ]−∞,−1[ and ]1,∞[.
  • f increases on ]−1,1[ and decreases on the other two intervals.
  • Example 5 Prove that f(x)=tan⁡x−x is increasing on ]0,π2[.
  • Solution:
  • f′(x)=sec⁡2x−1=tan⁡2x.
  • On ]0,π2[, tan⁡x>0, so f′(x)>0.
  • Hence f is increasing there.

Notes

  • The values that separate increasing from decreasing are those where f′(x)=0 or f′(x) does not exist.
  • A function can be strictly increasing even if f′ is zero at one isolated point, such as f(x)=x3 at x=0.
  • If f′<h′ everywhere, then z=f−h has z′=f′−h′<0 everywhere, so z is decreasing on R.
  • Graphing software confirms the table of signs the curve rises where the table says "+" and falls where it says "−".
Increasing, Decreasing Functions and Maximum and Minimum Values · 3 / 9
2-2 (a) Critical points

2-2 (a) Critical points

Definition

  • Let f be continuous on an interval and let c be inside it.
  • The point (c,f(c)) is a critical point if f′(c)=0, or f′(c) does not exist (for example a corner or a vertical tangent), while f is defined at c.
  • A critical point with f′(c)=0 is also called a stationary point.
  • At a hill top the tangent is horizontal and f changes from increasing to decreasing.
  • At a valley bottom the tangent is horizontal and f changes from decreasing to increasing.
  • At a sharp corner the left and right derivatives differ, so f′ does not exist, yet the point can still be a turning point.
Increasing, Decreasing Functions and Maximum and Minimum Values · 4 / 9
2-2 (b) Local maximum and minimum values

2-2 (b) Local maximum and minimum values

Definitions

  • f(c) is a local maximum value if there is an open interval around c in which f(x)≤f(c) for all x.
  • f(c) is a local minimum value if there is an open interval around c in which f(x)≥f(c) for all x.
  • Together they are called local extrema.
  • A local maximum can be smaller than a local minimum elsewhere on the graph, because the comparison is only with nearby points.

Necessary condition

  • If f has a local extremum at an inner point c, then f′(c)=0 or f′(c) does not exist.
  • The converse is false f(x)=x3 has f′(0)=0 but no extremum at 0.
Increasing, Decreasing Functions and Maximum: a wire curve y = x³ – 3x like a hill and valley shows curve behaviour and maximum and minimum values: max (-1, 2) and min (1, -2).
Increasing, Decreasing Functions and Maximum: a wire curve y = x3 – 3x like a hill and valley shows curve behaviour and maximum and minimum values: max (-1, 2) and min (1, -2).
Increasing, Decreasing Functions and Maximum→a wire curve y = x3 – 3x like a hill and valley shows curve behaviour and maximum and minimum values: max (-1, 2) and min (1, -2).: 2-2 (b) Local maximum and minimum values
Increasing, Decreasing Functions and Maximum and Minimum Values · 5 / 9
2-2 (c) First-derivative test for local extrema

2-2 (c) First-derivative test for local extrema

Rule

  • Let (c,f(c)) be a critical point of the continuous function f.
  • If f′ changes from positive to negative at c, then f(c) is a local maximum.
  • If f′ changes from negative to positive at c, then f(c) is a local minimum.
  • If f′ does not change sign at c, there is no local extremum.
  • Find the domain and f′(x).
  • Find the critical points (where f′=0 or does not exist, inside the domain).
  • Make a table of signs of f′ around each critical point.
  • Decide maximum, minimum or neither, and calculate f(c).

✍️ Worked examples

  • Example 6 Find the local extrema of f(x)=x3+3x2−9x−7.
  • Solution:
  • f′(x)=3x2+6x−9=3(x+3)(x−1).
  • Critical points at x=−3 and x=1.
  • Signs of f′ positive for x<−3, negative for −3<x<1, positive for x>1.
  • At x=−3 the sign goes +→− local maximum f(−3)=−27+27+27−7=20.
  • At x=1 the sign goes −→+ local minimum f(1)=1+3−9−7=−12.
  • Example 7 Find the local extrema of f(x)=x2/3(2x−5).
  • Solution:
  • f(x)=2x5/3−5x2/3, defined and continuous on R.
  • f′(x)=103x2/3−103x−1/3=103⋅x−1x1/3.
  • f′(0) does not exist and f′(1)=0, so the critical points are x=0 and x=1.
  • For x<0 x−1<0 and x1/3<0, so f′>0. For 0<x<1: f′<0. For x>1: f′>0.
  • At x=0 the sign goes +→− local maximum f(0)=0 (at a sharp point where the derivative does not exist).
  • At x=1 the sign goes −→+ local minimum f(1)=−3.
  • Example 8 Find the local extrema of f(x)=x+4x.
  • Solution:
  • The domain is R−{0}.
  • f′(x)=1−4x2=(x−2)(x+2)x2.
  • Critical points x=±2 (the value x=0 is not in the domain).
  • Signs positive for x<−2, negative on ]−2,0[ and on ]0,2[, positive for x>2.
  • Local maximum f(−2)=−4 and local minimum f(2)=4.
  • Notice that the local maximum −4 is less than the local minimum 4.
  • Example 9 Show that f(x)=x3+3x−4 has no local extrema.
  • Solution:
  • f′(x)=3x2+3>0 for every x.
  • There is no critical point and f is always increasing, so no local maximum or minimum exists.
  • Example 10 Find the local extrema of f(x)=xe−x.
  • Solution:
  • f′(x)=e−x−xe−x=(1−x)e−x.
  • The factor e−x is always positive, so f′=0 only at x=1.
  • f′>0 for x<1 and f′<0 for x>1, so f(1)=e−1=1e is a local maximum.
Increasing, Decreasing Functions and Maximum and Minimum Values · 6 / 9
2-2 (d) Absolute maximum and minimum on a closed interval

2-2 (d) Absolute maximum and minimum on a closed interval

Definition

  • Let f be defined on [a,b] and c∈[a,b].
  • f(c) is the absolute minimum if f(c)≤f(x) for all x∈[a,b].
  • f(c) is the absolute maximum if f(c)≥f(x) for all x∈[a,b].
  • Together they are the absolute extrema (extreme values). They may occur at an inner point or at an end point.
  • If f is continuous on [a,b], it always has both an absolute maximum and an absolute minimum there.
  • For an open or unbounded interval, or a discontinuous function, they may not exist.
  • Find f(a) and f(b).
  • Find the critical points in ]a,b[ and the value of f at each.
  • Compare all these values the largest is the absolute maximum and the smallest is the absolute minimum.

✍️ Worked examples

  • Example 11 Find the absolute extrema of f(x)=x3−12x+12 on [−3,3].
  • Solution:
  • f(−3)=−27+36+12=21 and f(3)=27−36+12=3.
  • f′(x)=3x2−12=3(x−2)(x+2), so the critical points are x=±2, both inside the interval.
  • f(2)=8−24+12=−4 and f(−2)=−8+24+12=28.
  • Compare 21,3,−4,28 absolute maximum =28 at x=−2, absolute minimum =−4 at x=2.
  • Example 12 Find the absolute extrema of f(x)=xe−x on [0,2].
  • Solution:
  • f′(x)=(1−x)e−x, critical point x=1.
  • f(0)=0, f(1)=1e≈0.368, f(2)=2e2≈0.271.
  • Absolute maximum 1e at x=1; absolute minimum 0 at the end point x=0.
  • Example 13 Find the absolute extrema of f(x)=sin⁡x+cos⁡x on [0,2π].
  • Solution:
  • f′(x)=cos⁡x−sin⁡x=0 gives tan⁡x=1, so x=π4 or 5π4.
  • f(π4)=2 and f(5π4)=−2.
  • End points f(0)=f(2π)=1.
  • Absolute maximum 2 and absolute minimum −2.
  • Example 14 Find the absolute extrema of f(x)=x−1 on [2,5].
  • Solution:
  • f′(x)=12x−1>0, so there is no critical point and f is increasing.
  • The extrema are at the end points minimum f(2)=1 and maximum f(5)=2.

Remarks

  • On a closed interval the end points must always be tested, because an extreme value can sit there with a non-zero derivative.
  • A function may have several local extrema but only one absolute maximum value (it can be attained at more than one point).
  • A local extremum is only "local"; the absolute extremum of an interval can coincide with a local extremum.
Increasing, Decreasing Functions and Maximum and Minimum Values · 7 / 9
2-2 (e) Finding unknown constants from extrema

2-2 (e) Finding unknown constants from extrema

Idea

  • Use the facts "the point lies on the curve" and "f′(c)=0 at the extremum" to build equations.

✍️ Worked examples

Comparison aspectMain ideaKeywords
Example 15The curve f(x)=ax3+bx2 has a local maximum at (1,4). Find a and b.curvelocalmaximumFind
•SolutionSolution
The point lies on the curvea+b=4
Critical point
  • f′(x)=3ax2+2bx
  • so f′(1)=3a+2b=0
3ax2
•From the first equation b=4−a; then 3a+8−2a=0, so a=−8 and b=12.firstequationthen
Check
  • f′(x)=−24x2+24x=24x(1−x) is positive just before x=1 and negative just after
  • so x=1 is indeed a maximum
24x2positivejustbefore
Increasing, Decreasing Functions and Maximum and Minimum Values · 8 / 9
2-2 (f) Summary

2-2 (f) Summary

Point of comparison Monotony test Local extremum test Absolute extremum on [a,b]
Uses Sign of f′ on intervals Sign change of f′ at a critical point Values of f at critical points and end points
Result Increasing / decreasing intervals Local maximum / minimum value Greatest and least value on the interval
Key condition f differentiable on the interval f continuous at c f continuous on [a,b]
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Increasing, Decreasing Functions and Maximum and Minimum Values · 9 / 9