Lesson 15 Concavity and the Second Derivative Test

LESSON 15 CONCAVITY AND THE SECOND DERIVATIVE TEST

Definition Let f be a function that is differentiable at c. Then

1.the graph of f is concave upward at the point if there exists an open interval containing c such that on the interval the graph of f is above the tangent line through the point P,

2.the graph of f is concave downward at the point if there exists an open interval containing c such that on the interval the graph of f is below the tangent line through the point P.

Theorem (Test for Concavity) Suppose that a function f is differentiable on an open interval containing c and that exists. Then

1.if , then the graph of f is concave upward at the point ,

2.if , then the graph of f is concave downward at the point .

Definition A point on the graph of a function at which the concavity changes is called a point of inflection or an inflection point.

Example Find the interval(s) on which the following functions are concave upward and concave downward. Also, find the point(s) of inflection ( or inflection point(s).)

Sign of : +



NOTE: Since when x is in the interval , then the graph of the function f is concave upward on this interval. Since when x is in the interval , then the graph of the function f is concave downward on this interval.

NOTE: Since is in the domain of the function f, then an inflection point occurs when . Now, we will calculate the y-coordinate of this point:

= =

= = =

Answer:Concave Upward:

Concave Downward:

Inflection Point(s):

NOTE:

Sign of : + +



NOTE: Since when x is in the set , then the graph of the function y is concave upward on this set. Since when x is in the set , then the graph of the function y is concave downward on this set.

NOTE: Since the numbers , 0, and are in the domain of the

function y, then inflection points occurs when , , and

. Now, we will calculate the y-coordinate of these points:

For : =

NOTE:

For : =

For :

Answer:Concave Upward:

Concave Downward:

Inflection Point(s): ; ;

Sign of : + +



NOTE: Since when t is in the set and the number 0 is in the domain of the function s, then the graph of the function s is concave upward on the interval . Since when t is in the interval , then the graph of the function s is concave downward on this interval.

NOTE: Since is in the domain of the function f, then an inflection point occurs when . Now, we will calculate the y-coordinate of this point:

Using a calculator, we have that

= = 2132

Not using a calculator, we have that

= =

= = =

= = 2132

Answer:Concave Upward:

Concave Downward:

Inflection Point(s):

NOTE: In order to obtain the second derivative of the function y, you could use the Product Rule. However, using properties of exponents, we have that

= =

Thus, = . Using the Chain Rule

again, we have that

= =

NOTE: and

Sign of : + +



0 4

NOTE: Since when u is in the set , then

the graph of the function y is concave upward on this set. Since

when u is in the set , then the graph of the function y is concave downward on this set.

NOTE: Since the numbers 0, , and 4 are in the domain of the function y, then inflection points occurs when , , and . Now, we will calculate the y-coordinate of these points:

For :

For : =

For :

Answer:Concave Upward:

Concave Downward:

Inflection Point(s): ; ;

NOTE: The domain of the function f is the interval .

= =

NOTE: . However, is not in the domain of the

function f .

Sign of :



NOTE: Since when x is in the interval , then the graph of the function f is concave downward on this interval. Since is not positive anywhere, then the graph of the function f is not concave upward anywhere.

Answer:Concave Upward: Nowhere

Concave Downward:

Theorem (The Second Derivative Test for Local Maximums and Minimums) Let c is a critical number of a function f. If f is differentiable on an open interval containing c, then

1.if , then f has a local maximum occurs at and the local maximum is ,

2.if , then f has a local minimum occurs at and the local minimum is ,

3.if , then the test fails and you will need to use the First Derivative Test to classify whether a local maximum, local minimum, or neither occurs at .

NOTE: You would use the Second Derivative Test if it was the second derivative of the function can be calculate quickly and you didn’t want the information about when the function is increasing and decreasing. Remember, if you have the sign of the first derivative, then the First Derivative Test will give you the information about local maximums and local minimums.

Examples Find the critical numbers for the following functions. If the second derivative of the following functions can be calculated quickly, then find the local maximum(s) and local minimum(s) using the Second Derivative Test. If the second derivative can not be calculated quickly, then state this and do not find the local maximum(s) and local minimum(s).

NOTE: The domain of the function f is the set of all real numbers.

Critical Number: 3

loc max occurs at

Answer:Local Maximum(s):

Local Minimum(s): None

NOTE: The domain of the function y is the set of all real numbers such that and .

We differentiated this function in Lesson 9 obtaining that

Critical Number:

The second derivative can not be calculated quickly. It would be faster to use the First Derivative Test to classify the local maximum(s) and local minimum(s). See this work in Lesson 14.

NOTE: The domain of the function g is the set of all real numbers such that and .

We differentiated this function in Lesson 9 obtaining that

Critical Numbers: 0 and

The second derivative can not be calculated quickly. It would be faster to use the First Derivative Test to classify the local maximum(s) and local minimum(s). See this work in Lesson 14.

NOTE: The domain of the function y is the set of all real numbers such that and .

We differentiated this function in Lesson 9 obtaining that

Critical Numbers: , 5

The second derivative can not be calculated quickly. It would be faster to use the First Derivative Test to classify the local maximum(s) and local minimum(s). See this work in Lesson 14.

NOTE: The domain of the function h is the set of all real numbers.

We differentiated this function in Lesson 10 obtaining that

Critical Number(s): , , 3

The second derivative can not be calculated quickly. It would be faster to use the First Derivative Test to classify the local maximum(s) and local minimum(s). See this work in Lesson 14.

NOTE: The domain of the function s is the set of all real numbers.

= =

Critical Number(s): , 5

loc max occurs at

loc min occurs at

= =

Answer: Local Maximum(s):

Local Minimum(s):

NOTE: The domain of the function y is the set of all real numbers.

Critical Numbers: , 0

: loc min occurs at

: the test fails

Using the First Derivative Test, we have that

Inc Inc

Sign of : X + +



Thus, neither a local maximum nor a local minimum occurs at since the function y is increasing on the left-hand side of and it is still increasing on the right-hand side of .

NOTE: We do not need the sign of on the interval in order to classify what is happening at the critical number of .

Using a calculator: or

= = =

Answer: Local Maximum(s): None

Local Minimum(s):

NOTE: The domain of the function g is the set of all real numbers such that .

= =

Critical Numbers:

loc max occurs at

Since , then =

= = =

Answer: Local Maximum(s):

Local Minimum(s): None

NOTE: The domain of the function f is the interval .

= =

Critical Number(s): 3

loc min occurs at

Since , then

Answer: Local Maximum(s): None

Local Minimum(s):