Monotonicity and Concavity在线视频

Today we study the Calculus (I)

The topic is Monotonicity and Concavity

Now look at the question

How to determine

the monotonicity and concavity of some function

Our idea

is by the signs of the first derivative

and the second derivative of our function

Pay attention to the signs

What signs

Yes

the first derivative means f

the first prime

positive or negative

The second derivative may be positive

may be negative

means double prime

Now look at the definition

Let f be defined on an interval I

The I may be open interval

may be closed interval

and may be neither

We will say that Case 1

f is increasing function on I if

for every pair of number x1 and x2

If x1 less than x2

we can get

the corresponding two values less than too

At this time

the corresponding variables less than

values less than

means f is our increasing function

Case 2

f is decreasing on I if

for every pair of numbers x1 and x2

Corresponding

the two variables are less than

and the corresponding values

large than

f(x1) larger than f(x2)

At this time

f is decreasing function

Together

Case 3

f is strictly monotonic on the given interval I if

it is either increasing on I or decreasing on I

Remember increasing or decreasing

or strictly monotonic

Now look at the Theorem of Monotonicity

Theorem A

Let f be continuous function on an interval I

and differentiable at every interior point of I

Case 1

if f prime x is larger than 0

for all x interior to I

and then

f is increasing on I

Why

From the sign of the first derivative

f prime larger than 0

and f is increasing

Look at the figure

f corresponding prime large than 0

Look at Case 2

if f prime x less than 0

for all x interior to I

and then f is decreasing on I

From this figure

we can find

f prime less than 0

So remember prime larger than 0

means increasing

corresponding

f prime less than 0

corresponding decreasing function

Look at Definition of Concavity

Let f be differentiable

on an open interval I

We will say that f is concave up on I

if f prime is increasing on I

f prime

pay attention to not f

it is f prime increasing

at this time concave up

Corresponding

we will say that

f is concave down on I

if f prime is decreasing

means decreasing concave down

Try to understand the definition

Look at Theorem B

let f be twice differentiable on the open interval I

At this time

if f the second derivative

is larger than 0

for all x in I

and then f is concave up

f double prime larger than 0

corresponding concave up

Look at the figure

f prime is increasing

corresponding

double prime larger than 0

So from the Figure 1

try to understand the definition

In the similar way

we give another definition

f the second prime less than 0 for all x in I

And then

f is concave down on I

From the figure

we can find f prime x decreasing

corresponding double prime less than 0

So it is our

concave down and concave up

Larger than 0

concave up

less than 0

concave down

Try to use Theorem A and Theorem B

to do our Examples

Look at Example 1

If y equals e x power minus x minus 1

try to find

where f is increasing

means what numbers for x

f is increasing

and where f is decreasing

How to do

Try to find the corresponding first derivative signs

Now look at

the corresponding domain is

the real number system

means from minus infinity to positive infinity

y prime

y is given

corresponding prime

equals e x minus 1

What signs

Larger than 0 or less than 0

So we should know

on what values y prime equals 0

Because in this corresponding domain

y prime less than 0

y prime equals 0

means e 0 power minus 1 equals 0

So x less than 0

corresponding y prime less than 0

In the similar way

larger than 0

y prime larger than 0

So the function is

decreasing on minus infinity to 0

Pay attention to the 0

closed means our end point

Corresponding

increasing on

from 0 to positive infinity

Of course including the 0 point

Pay attention to

consider the signs of y prime

We will write the open interval

from minus infinity to 0

0 to positive infinity

Consider the sign of prime

If we give the conclusion

we should write

including the point

means closed point

So now look at Example 2

If f equals 2 x cube

minus 9 x square plus 12 x minus 3

try to find where f is increasing

and where it is decreasing

The similar idea

in Example 1

so try to find our domain

Yes it is a polynomial

So the domain is the real number system

means from minus infinity to positive infinity

f prime

f prime equals 6 x square minus 18 x plus 12

We can get the solution

f prime x write it in this form

6 time x minus 1

time x minus 2

By solving the equation

f prime x equals 0

We can get the two solutions

One point is 1

the other point is 2

Together from this

minus infinity to positive infinity

There are two special points

1 point and 2 point

In this corresponding domain

positive sign

This sign is minus

This sign is positive

So the domain is split

One two three

one two three

three domains

We can get the table

x between minus infinity to 1

and x between 1 and 2

x between the third domain

from 2 to positive infinity

We can get

the corresponding signs of f prime

Why

Because we have taken out

x equals 1

x equals 2

equals 0

Otherwise may be positive

may be negative

So we can get

the f prime signs

positive

negative and positive

If f prime is positive

according to our definition

corresponding

f increasing

corresponding decreasing

corresponding increasing

So we can get the following conclusion

The function is decreasing on

decreasing on this

In the open interval from 1 to 2

f prime negative

It is decreasing

Our solution is the function is

decreasing on closed

from 1 to 2

corresponding

increasing on minus infinity to 1

close the point

and 2 to positive infinity

close the 2 point

So now we look at Example 3

If y equals x cube

try to find where y is concave up

The new definition concave up

And where

it is concave down

concave down

Do you remember

definition concave up and concave down

Yes

f double prime

so now we know

the corresponding figure y equals x cube

Where

Yes in the red curve and blue curve

how to understand this

y is given

the first derivative

y prime equals 3 x square

y the second derivative equals 6 x

We consider y double prime

So

x less than 0

at this time

y double prime less than 0 too

So y is concave down on the corresponding domain

from minus infinity to 0

close the end point

Similarly

x is larger than 0

At this time

y double prime is larger than 0 too

So according to our definition

y is concave up

on the closed 0 to positive infinity

So pay attention to the point (0, 0)

it is the special point

The point (0, 0) is our

demarcation point from concaving up to concaving down

So we should consider this point

ok special point

So now we have our summary

The Monotonicity Theorem

let f be continuous on an interval I

and differentiable at every interior point of I

Case 1

f prime larger than 0 for all x

and then f is

Yes increasing on I

Corresponding

if f prime x less than 0 for all x

then x less than 0 means

f is decreasing on interval I

How about our Concavity Theorem

Let f be twice differentiable

on the open interval I

Twice means we can get

the second derivative of function f

Case 1

f the second derivative larger than 0

at this time

f is concave up on the given I

Corresponding

f the second derivative less than 0

f is concave down on I

Try to understand larger than 0

increasing

The first sign less than 0

decreasing

The second derivative

larger than 0 concave up

less than 0 concave down

Using the summary

try to finish our Question 1

Look at Question 1

If f(x) equals

double x cube minus 3 x square minus 12 x plus 7

find where f is increasing

and where it is decreasing

Do you remember the four steps

How to write it

Yes

f prime x equals this form

We try to find the points

f prime x equals 0

So we write it in this form

equals 6 time x minus 2 time x plus 1

d then we can find

the split points are minus 1 and 2

Corresponding domain

is split into 3 domains

First

from minus infinity to minus 1

Second

between minus 1 and 2

The third

from 2 to the positive infinity

We determine the signs

the corresponding different intervals

In this interval

the sign is positive

Between minus 1 and 2

negative

Another is positive

So we can make up the table

x between the 3 different intervals

f prime x

positive

negative

positive

According to our definition

we can determine

f(x) decreasing or increasing

f(x) increasing

decreasing

increasing

So we can get the corresponding conclusion

f(x) is increasing on

this interval and this interval

including the end points

means on minus infinity to minus 1

close the point

and from 2

include the point

to positive infinity

And f(x) is decreasing on minus 1 and 2

Now look at Question 2

If g(x) is given

try to find where g is increasing

and where g is decreasing

g prime x

try to get

Yes

g is u over v

u prime v minus u v prime over v square

So we can get corresponding g prime

It is our

the first derivative

From the denominator is positive at everywhere

ok

Determine the signs

we will consider the numerator

Numerator equals 0 means

corresponding split points are minus 1 and 1

And then we can get corresponding

this 3 corresponding intervals

minus 1 and 1

Determine the signs are negative

positive and negative

And then

we can get our conclusion

g(x) is increasing on closed interval minus 1 and 1

And then

g(x) is decreasing on

minus infinity to minus 1

and from 1 to positive infinity

So try to understand the question

The first step

get the g prime means the first derivative

Try to find the corresponding split points

How many split points

Maybe n

so the corresponding original interval

is divided into n plus 1 intervals

And then determine

which is our increasing interval

Which is our decreasing interval

Try to understand the question

So now class is over

See you next time