Exploring Concavity and Second Derivative in Calculus

MAT 1221 Survey of Calculus Section 3.3 Concavity and the Second Derivative Test http://myhome.spu.edu/lauw

Expectations Check your algebra. Check your calculator works Formally answer the question with the expected information

1 Minute You can learn all the important concepts in 1 minute.

1 Minute Critical numbers give the potential local max/mins

1 Minute Critical numbers give the potential local max/mins If the graph is concave down at a critical number, it has a local max

1 Minute Critical numbers give the potential local max/mins If the graph is concave up at a critical number, it has a local min

1 Minute You can learn all the important concepts in 1 minute. We are going to develop the theory carefully so that it works for all the functions that we are interested in. There are a few definitions

Preview Second Derivative Concavities Find the intervals of concave up and concave down The Second Derivative Test Find local max and mins. The two processes are related but not the same.

Second Derivative

Second Derivative ( ) f x = + 5 3 2 x x d dx ( ) ( ) x = + 4 2 5 6 f x x d dx ( ) ( ) x = + 3 20 12 f x x

Higher Derivatives Given a function ?(?) = ( ) the = derivative of ( ) f x f x the first derivative of ( ) f x which is a function.

Higher Derivatives Given a function ? (?) d ( ) = ( ) ( ) f x f x dx the = derivative of ( ) f x the = second derivative of ( ) f x

Concave Up (a) A function ? is called concave upward on an interval ? if the graph of ? lies above all of its tangents on ?. (b) A function ? is called concave downward on an interval ? if the graph of ? lies below all of its tangents on ?.

Concavity ? is concave up on ? Potential local min.

Concavity ? is concave down on ? Potential local max.

Concavity y Concave down Concave up x c ? has no local max. or min. ? has an inflection point at ? = ?

Definition An inflection point is a point where the concavity changes (from up to down or from down to up)

Concavity Test Concavity Test (a) If ? (?) > 0 on an interval ?, then ? is concave upward on ?. (b) If ? (?) < 0 on an interval ?, then f is concave downward on ?.

Concavity Test Concavity Test (a) If ? (?) > 0 on an interval ?, then ? is concave upward on ?. (b) If ? (?) < 0 on an interval ?, then f is concave downward on ?. d dx ( ) Why? (Hint: ) f ( ) x ( ) = f x

Why? ? (?) > 0 implies ? (?) is increasing. i.e. the slope of tangent lines is increasing. d dx ( ) ( ) x ( ) = f f x

Why? ? (?) < 0 implies ? (?) is decreasing. i.e. the slope of tangent lines is decreasing. d dx ( ) ( ) x ( ) = f f x

Example 1 Find the intervals of concavity and the inflection points + = x x f + 3 2 ( ) 2 6 3 1 x x

= + + 3 2 ( ) 2 6 3 1 f x x x x Example 1 f f (x ) (x ) 1. Find , and the values of such that x x = ( ) 0 f

= + + 3 2 ( ) 2 6 3 1 f x x x x Example 1 2. Sketch a diagram of the subintervals formed by the values found in step 1. Make sure you label the subintervals.

= + + 3 2 ( ) 2 6 3 1 f x x x x Example 1 3. Find the intervals of concavity and inflection point. ( ) 1 f = 8

= + + 3 2 ( ) 2 6 3 1 f x x x x Example 1

Transition... The last process is to find the interval of concave up and down The next process is to find local max and min.

The Second Derivative Test The Second Derivative Test Suppose ? is continuous near ?. (a) If ? (?) = 0 and ? (?) > 0, then ? has a local minimum at c. (b) If ? (?) = 0 and ? (?) < 0, then f has a local maximum at ?. (c) If ? (?) = 0, then no conclusion (use 1st derivative test)

Second Derivative Test c = Suppose If then ? has a local min at ? = ? 0 ) ( c f ( ) 0 f y ? (?) > 0 ? (?) = 0 x c

Second Derivative Test c = Suppose If then ? has a local max at ? = ? 0 ) ( c f ( ) 0 f y ? (?) < 0 ? (?) = 0 x c

The Second Derivative Test The Second Derivative Test (c) If ? (?) = 0, then no conclusion

The Second Derivative Test The Second Derivative Test If ? (?) = 0, then no conclusion = = 4 ( ) ( ) f x x = = 3 4 0 0 f x x x = = 2 ( ) 12 (0) f x x = 2 12 0 0 f

The Second Derivative Test The Second Derivative Test If ? (?) = 0, then no conclusion = = 4 ( ) ( ) g x x = = 3 4 0 0 g x x x g x = = 2 ( ) (0) 12 12 0 x = 2 0 g

The Second Derivative Test The Second Derivative Test If ? (?) = 0, then no conclusion = = 3 ( ) ( ) h x x = = 2 3 0 0 h x x x = = = ( ) (0) 6 6 0 h x h x 0

The Second Derivative Test The Second Derivative Test Suppose ? is continuous near ?. (a) If ? (?) = 0 and ? (?) > 0, then ? has a local minimum at c. (b) If ? (?) = 0 and ? (?) < 0, then f has a local maximum at ?. (c) If ? (?) = 0, then no conclusion (use 1st derivative test)

Example 2 Use the second derivative test to find the local max. and local min. 9 ) ( = x x f + 3 2 24 10 x x

Example 2 (a) Find the critical numbers of ) ( = x x f + 3 2 9 24 10 x x

Example 2 (b) Use the Second Derivative Test to find the local max/min of = + 3 2 ( ) 9 24 10 f x x x x The local max. value of ? is The local min. value of ? is ( ) 2 ( ) 4 = = 10, 6 f f

Review Example 1 & 2 illustrate two different but related problems. 1. Find the intervals of concavity and inflection points. 2. Find the local max. /min. values

Expectations Follow the steps to solve the two problems