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
Increasing
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 is increasing on an interval if its graph rises from left to right: gives .
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 -axis, so its slope is positive.
Where it decreases, the tangent makes an obtuse angle, so is negative.
Where the tangent is horizontal, .
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 be differentiable on the open interval .
If for every , then is increasing on .
If for every , then is decreasing on .
Find the domain of and check where it is continuous and differentiable.
Find .
Solve and find where does not exist; these values separate the intervals.
Make a table of signs of 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 .
Solution:
is a polynomial, so it is continuous and differentiable on .
.
at and .
Sign of f′ positive on ]−∞,−1[, negative on ]−1,1[, positive on ]1,∞[.
So is increasing on and on , and decreasing on .
Example 2 Find the intervals of monotony of for .
Solution:
.
gives , so or .
Test points at x=π2, f′=1>0; at x=π, f′=−1<0; at x=3π2, f′=1>0.
Increasing on and ; decreasing on .
Example 3 Find the intervals of monotony of .
Solution:
The domain is .
.
gives (since ).
For , ; for , .
is increasing on and decreasing on .
Example 4 Find the intervals of monotony of .
Solution:
.
The denominator is always positive, so the sign depends on .
on and on and .
increases on and decreases on the other two intervals.
Example 5 Prove that is increasing on .
Solution:
.
On , , so .
Hence is increasing there.
Notes
The values that separate increasing from decreasing are those where or does not exist.
A function can be strictly increasing even if is zero at one isolated point, such as at .
If everywhere, then has everywhere, so is decreasing on .
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 be continuous on an interval and let be inside it.
The point is a critical point if , or does not exist (for example a corner or a vertical tangent), while is defined at .
A critical point with is also called a stationary point.
At a hill top the tangent is horizontal and changes from increasing to decreasing.
At a valley bottom the tangent is horizontal and changes from decreasing to increasing.
At a sharp corner the left and right derivatives differ, so 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
is a local maximum value if there is an open interval around in which for all .
is a local minimum value if there is an open interval around in which for all .
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 has a local extremum at an inner point , then or 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).: 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 be a critical point of the continuous function .
If changes from positive to negative at , then is a local maximum.
If changes from negative to positive at , then is a local minimum.
If does not change sign at , there is no local extremum.
Find the domain and .
Find the critical points (where or does not exist, inside the domain).
Make a table of signs of around each critical point.
Decide maximum, minimum or neither, and calculate .
✍️ Worked examples
Example 6 Find the local extrema of .
Solution:
.
Critical points at and .
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 .
Solution:
, defined and continuous on .
.
does not exist and , so the critical points are and .
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 .
Solution:
The domain is .
.
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 and local minimum .
Notice that the local maximum is less than the local minimum .
Example 9 Show that has no local extrema.
Solution:
for every .
There is no critical point and is always increasing, so no local maximum or minimum exists.
Example 10 Find the local extrema of .
Solution:
.
The factor is always positive, so only at .
for and for , so 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 be defined on and .
is the absolute minimum if for all .
is the absolute maximum if for all .
Together they are the absolute extrema (extreme values). They may occur at an inner point or at an end point.
If is continuous on , 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 and .
Find the critical points in and the value of 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 on .
Solution:
and .
, so the critical points are , both inside the interval.
and .
Compare 21,3,−4,28 absolute maximum =28 at x=−2, absolute minimum =−4 at x=2.
Example 12 Find the absolute extrema of on .
Solution:
, critical point .
, , .
Absolute maximum at ; absolute minimum at the end point .
Example 13 Find the absolute extrema of on .
Solution:
gives , so or .
and .
End points f(0)=f(2π)=1.
Absolute maximum and absolute minimum .
Example 14 Find the absolute extrema of on .
Solution:
, so there is no critical point and 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 " at the extremum" to build equations.
✍️ Worked examples
Comparison aspect
Main idea
Keywords
Example 15
The curve has a local maximum at . Find and .
curvelocalmaximumFind
•
Solution
Solution
The point lies on the curve
a+b=4
Critical point
f′(x)=3ax2+2bx
so f′(1)=3a+2b=0
3ax2
•
From the first equation ; then , so and .
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