Transcript Increasing & Decreasing Functions & The 1st Derivative Test

Increasing & Decreasing
st
1
Functions & The
Test
Chapter 4.3
Derivative
Increasing & Decreasing Functions
• In the first section of this chapter, you learned how to find points at
which relative extrema occur
• But no theorem in that section told us how to determine whether a
relative extremum is a relative maximum or a relative minimum
• In this section you will learn, first, how the derivative can be used to
determine whether a function is increasing, decreasing, or constant on
an interval
• Second, you will learn how to determine whether a relative extremum
is a relative maximum or a relative minimum
Definitions of Increasing & Decreasing
Functions
DEFINITION:
A function f is increasing on an interval if for any two numbers 𝑥1 and 𝑥2 in the interval, 𝑥1 < 𝑥2 ⟹ 𝑓 𝑥1 <
𝑓(𝑥2 ).
A function f is decreasing on an interval if for any two numbers 𝑥1 and 𝑥2 in the interval, 𝑥1 < 𝑥2 ⟹ 𝑓 𝑥1 >
𝑓 𝑥2 .
In other words, as the values of x move from left to right, the function is decreasing if the function values
decrease; the function is increasing if the function values increase.
Definitions of Increasing & Decreasing
Functions
Definitions of Increasing & Decreasing
Functions
Definitions of Increasing & Decreasing
Functions
The Derivative & Increasing/Decreasing
Functions
• The next theorem will show that, on an interval where a function is
decreasing, the slopes of all tangent lines have negative values
• Where a function is increasing, the slopes of all tangent lines have
positive values
• Note that relative minima and relative maxima mark the points at
which a function changes from increasing to decreasing or vice versa
• When determining whether a function is increasing or decreasing, we
will first find the relative extrema and then pick points on either side
to check where the slopes of the tangent lines are positive and where
they are negative
The Derivative & Increasing/Decreasing
Functions
Theorem 4.5: Test for Increasing &
Decreasing Functions
THEOREM:
Let f be a function that is continuous on the closed interval [𝑎, 𝑏] and differentiable on the open interval (𝑎, 𝑏)
1. If 𝑓 ′ 𝑥 > 0 for all x in (𝑎, 𝑏), then f is increasing on [𝑎, 𝑏].
2. If 𝑓 ′ 𝑥 < 0 for all x in (𝑎, 𝑏), then f is decreasing on [𝑎, 𝑏].
3. If 𝑓 ′ 𝑥 = 0 for all x in (𝑎, 𝑏), then f is constant on [𝑎, 𝑏]
We will only prove #1. We must end the proof by concluding that if 𝑥1 < 𝑥2 , then 𝑓 𝑥1 < 𝑓(𝑥2 ), as per the
definition of an increasing function.
PROOF
Assume that 𝑓 ′ 𝑥 > 0 for all x in some interval [𝑥1 , 𝑥2 ].
Theorem 4.5: Test for Increasing &
Decreasing Functions
THEOREM:
Let f be a function that is continuous on the closed interval [𝑎, 𝑏] and differentiable on the open interval (𝑎, 𝑏)
1. If 𝑓 ′ 𝑥 > 0 for all x in (𝑎, 𝑏), then f is increasing on [𝑎, 𝑏].
PROOF
Assume that 𝑓 ′ 𝑥 > 0 for all x in some interval [𝑥1 , 𝑥2 ]. By the MVT there exists c in (𝑥1 , 𝑥2 ) such that
𝑓′ 𝑐 =
𝑓 𝑥2 − 𝑓 𝑥1
𝑥2 − 𝑥1
Since 𝑓 ′ 𝑐 > 0 and 𝑥2 − 𝑥1 > 0, then we must have that 𝑓 𝑥2 − 𝑓 𝑥1 > 0. But this means that
𝑥1 < 𝑥2 ⟹ 𝑓 𝑥1 < 𝑓(𝑥2 )
which is exactly the statement of the definition of an increasing function. Therefore, f is increasing on [𝑎, 𝑏].
Example 1: Intervals on Which f is Increasing
or Decreasing
Find the open intervals on which 𝑓 𝑥 =
decreasing.
𝑥3
−
3 2
𝑥
2
is increasing or
Relative extrema occur at critical numbers, but not all critical numbers occur at relative extrema (you’ll see an
example later). As a polynomial function, f is differentiable (and hence continuous) over all of ℝ so we can
apply theorem 4.5. Differentiate f and find the critical numbers
𝑓 ′ 𝑥 = 3𝑥 2 − 3𝑥
3𝑥 2 − 3𝑥 = 0 ⟹ 3𝑥 𝑥 − 1 = 0 ⟹ 𝑥 = 0 or 𝑥 = 1
We need to know what the tangent line slopes are like to the left and right of these points. Since, by theorem
4.2, relative extrema occur only at critical numbers, then we can choose any values of x and determine the value
of 𝑓′(𝑥) at these values. At 𝑥 = −1 we have 𝑓 ′ −1 = 3 −1 2 − 3 −1 = 6 > 0. So the function is
−3
increasing on the interval (−∞, 0). At 𝑥 = 0.5, 𝑓 ′ 0.5 = 3 0.5 2 − 3 0.5 = 4 < 0, so the function is
decreasing on the interval (0,1). At 𝑥 = 2, 𝑓 ′ 𝑥 = 3 2 2 − 3 2 = 6 > 0, so the function is increasing on the
interval (1, ∞).
Guidelines for Finding Intervals on Which a
Function is Increasing or Decreasing
Let f be continuous on the interval (𝑎, 𝑏). To find the open intervals on
which f is increasing or decreasing
1. Locate the critical numbers of f in (𝑎, 𝑏) and use these numbers to
determine test intervals
2. Determine the sign of 𝑓′(𝑥) at one test value in each of the intevals
3. Use Theorem 4.5 to determine whether f is increasing or decreasing
on each interval
These guidelines are also valid if the interval (𝑎, 𝑏) is replaced by an
interval of the form (−∞, 𝑏), (𝑎, ∞), or (−∞, ∞)
Guidelines for Finding Intervals on Which a
Function is Increasing or Decreasing
• As the final note in the guidelines suggest, a function may be
increasing or decreasing over its entire domain
• A function that is either increasing or decreasing over an interval or
over its entire domain is called strictly monotonic
• Would you expect to find critical numbers for a function that is strictly
monotonic?
Guidelines for Finding Intervals on Which a
Function is Increasing or Decreasing
The First Derivative Test
• Consider the results of Example 1
• Going from left to right, the function is increasing up to 0, then decreasing from 0 to
1
• Think of this as “ascending” to 0 and “descending” between 0 and 1
• That is, the shape of the graph is like a “hill”, so the critical number at 0 represents a
relative maximum
• The graph also “descends” from 0 to 1, then “ascends” after 1
• This is like a “valley” so we have a relative minimum
• However, the graph of 𝑓 𝑥 = 𝑥 3 has a horizontal tangent line at 0 but the
function does not change from increasing to decreasing
• The next theorem tells us how to use this to determine when we have a
relative maximum, a relative minimum, or neither
Theorem 4.6: The First Derivative Test
THEOREM:
Let c be a critical number of a function f that is continuous on an open interval I containing c. If f is
differentiable on the interval, except possibly at c, then 𝑓(𝑐) can be classified as follows
1. If 𝑓′(𝑥) changes from negative to positive at c, then f has a relative minimum at 𝑐, 𝑓 𝑐
2. If 𝑓′(𝑥) changes from positive to negative at c, then f has a relative maximum at 𝑐, 𝑓 𝑐
3. If 𝑓′(𝑥) is positive on both sides of c or negative on both sides of c, then 𝑓(𝑐) is neither a relative minimum
nor a relative maximum
The proof is simple. You should read it and understand why it is correct. Note that f need not be differentiable
at c. That is, if the derivative does not exist at c, we should still invoke this theorem to determine whether we
have relative maximum or a relative minimum.
Example 2: Applying the First Derivative Test
Find the relative extrema of the function 𝑓 𝑥 =
interval (0,2𝜋).
Find the critical numbers:
𝑓′ 𝑥 =
1
𝑥
2
− sin 𝑥 in the
1
− cos 𝑥
2
1
The derivative is zero where cos 𝑥 = 2. There are thus two critical numbers, one in the first quadrant and one in
the fourth quadrant. They are 𝑥 =
𝜋
𝜋
3
≈ 1.047 and 𝑥 =
1
5𝜋
3
1
≈ 5.236. At 𝑥 = 1, 𝑓 ′ 1 = 2 − cos 1 ≈ −0.040 <
0, so f is decreasing on 0, 3 . At 𝑥 = 2, 𝑓 ′ 2 = 2 − cos 2 ≈ 0.916 > 0, so f is increasing on
1
6, 𝑓 ′ 6 = 2 − cos 6 ≈ −0.460 < 0, so f is decreasing on
relative minimum at 𝑥 =
𝜋
3
and a relative maximum at 𝑥 =
5𝜋
, 2𝜋
3
5𝜋
.
3
𝜋 5𝜋
,
3 3
. At 𝑥 =
. Thus, by theorem 4.6 the function has a
Example 3: Applying the First Derivative Test
2
3
Find the relative extrema of 𝑓 𝑥 = 𝑥 2 − 4 .
Find the critical numbers:
2
𝑓 𝑥 = ⋅ 2𝑥 𝑥 2 − 4
3
′
−1
3
=
4𝑥
3
3 𝑥2 − 4
The critical numbers occur wherever the derivative is zero or where it is not differentiable. In this case, 𝑓 ′ 𝑥 =
0 at 𝑥 = 0 and f is not differentiable at 𝑥 = ±2. The intervals to check are
−∞, −2 , −2,0 , 0,2 , and (2, ∞)
If 𝑥 = −3, the derivative is 𝑓 ′ −3 =
If 𝑥 = −1, the derivative is 𝑓 ′ −1 =
interval.
4 −3
3
3 (−3)2 −4
4 −1
3
3
−1 2 −4
=
=
−12
3
3 5
< 0 so the function is decreasing on the first interval.
−4
3
−3 3
> 0 so the function is increasing on the second
Example 3: Applying the First Derivative Test
2
3
Find the relative extrema of 𝑓 𝑥 = 𝑥 2 − 4 .
If 𝑥 = 1, then 𝑓 ′ 1 =
If 𝑥 = 3, then 𝑓 ′ 3 =
4 1
3
=
3
=
3 12 −4
4 3
3 32 −4
4
3
−3 3
12
3
3 5
< 0, so the function is decreasing on the third interval.
> 0, so the function is decreasing on the fourth interval.
Now use theorem 4.6. Since 𝑓′ changes from decreasing to increasing at 𝑥 = −2, we have a relative minimum.
Since 𝑓′ changes from increasing to decreasing at 𝑥 = 0, we have a relative maximum. Since 𝑓′ changes from
decreasing to increasing at 𝑥 = 2, we have a relative minimum.
Consider the Domain of the Function
As the next example will show,
you must consider the domain of
the function because values of x
where the function is not defined
must be used with the critical
numbers to determine the test
intervals
Example 4: Applying the First Derivative Test
Find the relative extrema of 𝑓 𝑥 =
𝑥 4 +1
.
𝑥2
Because the function is not defined at 𝑥 = 0, we must use this value along with the critical numbers to find the
relative extrema.
4−2
2𝑥
𝑓′ 𝑥 =
𝑥3
This is equal to zero at 𝑥 = ±1. So the intervals are
−∞, −1 , −1,0 , 0,1 , (1, ∞)
Using 𝑥 = −2, 𝑓 ′ −2 =
4 −2
2 −2
−2 3
30
1
1
= −8 < 0, so f is decreasing. Using 𝑥 = − 2, 𝑓 ′ − 2 =
1 4
2 −2
−2
1 3
−2
15
= − −1 >
0, so f is increasing. Thus, by the First Derivative Test, f has a relative minimum at 𝑥 = −1. Using 𝑥 = 0.5,
2 24 −2
2 0.5 4 −2
15
′
′
𝑓 𝑥 =
=
−15
<
0,
so
f
is
decreasing.
Using
𝑥
=
2,
𝑓
2
=
=
> 0, so f is increasing.
.053
23
4
Thus, by the First Derivative Test, f has a relative minimum at 𝑥 = 1.
Example 5: The Path of a Projectile
Neglecting air resistance, the path of a projective that is propelled at an
angle 𝜃 is
𝑔 sec 2 𝜃 2
𝑦=
𝑥 + 𝑥 tan 𝜃 + ℎ
2
2𝑣0
where y is the height, x is the horizontal distance, g is acceleration due
to gravity, 𝑣0 is the initial velocity, and h is the initial height. Let 𝑔 =
− 32 feet per second per second, 𝑣0 = 24 feet per second, and ℎ = 9
feet. What is the value of 𝜃 that will produce a maximum horizontal
distance?
Example 5: The Path of a Projectile
Before beginning the solution, think about what this problem is asking for: we must find the angle 𝜃 that
𝑑𝑥
maximum value of x. That is, we are looking for 𝑑𝜃 = 0 such that the value of 𝜃 produces a relative maximum.
It is important not to plunge ahead and start trying to find a derivative because the height y is not important
(except that we want it to be zero). Hence, the equation we want includes only the variables x and 𝜃, with 𝜃 as
the independent variable. We must differentiate
32 sec 2 𝜃 2
− sec 2 𝜃 2
−
𝑥 + 𝑥 tan 𝜃 + 9 = 0 ⟺
𝑥 + 𝑥 tan 𝜃 + 9 = 0
2 24 2
36
We could use implicit differentiation, but we would then need a value for both x and 𝜃, so it is easier to solve for
the dependent variable in terms of the independent variable (i.e., for x in terms of 𝜃). But the equation above is
a quadratic, so we can solve for x using the quadratic formula.
𝑥=
− tan 𝜃 ±
sec 2 𝜃
− 4 ⋅ 9 ⋅ − 36
− sec 2 𝜃
2 ⋅ 36
tan2 𝜃
Example 5: The Path of a Projectile
𝑥=
𝑥=
− tan 𝜃 ±
tan2 𝜃
− tan 𝜃 ±
− sec 2 𝜃
18
sec 2 𝜃
− 4 ⋅ 9 ⋅ − 36
− sec 2 𝜃
2 ⋅ 36
tan2 𝜃
+ sec 2 𝜃
18 sin2 𝜃 + 1
18 tan 𝜃 ±
cos 𝜃
=
1
cos 2 𝜃
2 𝜃 sin2 𝜃 + 1
18
cos
𝑥 = 18 cos2 𝜃 tan 𝜃 ±
= 18 cos 𝜃 sin 𝜃 ± 18 cos 𝜃 sin2 𝜃 + 1
cos 𝜃
𝑥 = 18 cos 𝜃 sin 𝜃 ± sin2 𝜃 + 1
Example 5: The Path of a Projectile
𝑥 = 18 cos 𝜃 sin 𝜃 ± sin2 𝜃 + 1
We choose the positive so that 𝑥 ≥ 0 and differentiate with respect to 𝜃. In this case, we will use technology to
differentiate and find the value of 𝜃 at which the maximum value ocurs. We get 𝜃 ≈ 0.61548 radians, which is
about 35.3°. If we use this in the original equation (with 𝑦 = 0) and solve for x, the distance is approximately
25.5 feet.
Exercise 4.3a
• Page 226, #1-16, 18-72 multiples of 3
Exercise 4.3b
• Page 227, #73-78, 81-95 odds