For more information, please see full course syllabus of College Calculus: Level I

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For more information, please see full course syllabus of College Calculus: Level I

For more information, please see full course syllabus of College Calculus: Level I

### First Derivative Test, Second Derivative Test

- The tests are two options for determining whether a critical point is a local minimum, a local maximum, or neither.
- In general the Second Derivative test is easier to use.
- Note that the Second Derivative test can be inconclusive; in this case, switch to the First Derivative Test.
- Understanding the geometry behind these tests makes them easy to remember and use!

### First Derivative Test, Second Derivative Test

Find the points of inflection of f(x) = [1/2] x

^{3}− 3x^{2}+ 7x + 11- f′(x) = [3/2] x
^{2}− 6x + 7 - f"(x) = 3x − 6
- 3x − 6 = 0
- 3x = 6
- x = 2

There is a point of inflection at x = 2

Find the points of inflection of f(x) = [1/12] x

^{4}− [1/2] x^{2}+ 12- f′(x) = [1/3] x
^{3}− x - f"(x) = x
^{2}− 1 - x
^{2}− 1 = 0 - (x + 1)(x − 1) = 0
- x = −1, 1

The points of inflection are x = −1 and x = 1

Determine the concavity of f(x) = sinx on the interval (0, π)

- f′(x) = cosx
- f"(x) = −sinx
- −sinx = 0
- sinx = 0
- x = 0, π
- These two points of inflection are at the end of the interval meaning everything in between must have the same concavity.
- f"([(π)/2]) = −sin[(π)/2] = −1
- f"(x) < 0 on the interval

f(x) = sinx is concave down on the interval (0, π)

Determine the concavity of f(x) = e

^{x}- f′(x) = e
^{x} - f"(x) = e
^{x} - There are no points of inflection, e
^{x}> 0

f(x) = e

^{x}is concave up.Determine the concavity of f(x) = 3x

^{2}−7x + 21- f′(x) = 6x − 7
- f"(x) = 6
- No points of inflection and f"(x) > 0

f(x) = 3x

^{2}− 7x + 21 is concave up.Determine the concavity of f(x) = x

^{3}− 12x- f′(x) = 3x
^{2}− 12 - f"(x) = 6x
- 6x = 0
- x = 0
- Inflection point at x = 0
- f"(x) < 0 for x < 0
- f"(x) > 0 for x > 0

f(x) = x

^{3}− 12x is concave down on x < 0 and concave up on x > 0Find any critical points of f(x) = x

^{3}− 12x and determine whether they represent local minimums or maximums.- f′(x) = 3x
^{2}− 12 - 3x
^{2}− 12 = 0 - x
^{2}− 4 = 0 - x
^{2}= 4 - x = ±2
- The critical points are x = −2 and x = 2
- From the previous problem we know that the function is concave down at x = −2 and concave up at x = 2. This tells us that x = −2 is a local maximum and x = 2 is a local minimum.

Local maximum at x = −2 and a local minimum at x = 2

Find the critical points and the concavity of f(x) = lnx

- f′(x) = [1/x]
- f′(x) is undefined at x = 0. Critical point at x = 0
- f"(x) = −[1/(x
^{2})] - No change in concavity, f"(x) never reaches zero.
- f"(x) < 0

f(x) = lnx is concave down and has a critical point at x = 0

Determine the concavity of f(x) = x

^{[1/3]}- f′(x) = [1/3] x
^{−[2/3]} - f"(x) = −[2/9] x
^{−[5/3]} - f"(0) and f′(0) are undefined and f"(x) ≠ 0.
- For x < 0, f"(x) > 0
- For x > 0, f"(x) < 0

f(x) = x

^{[1/3]}is concave up when x < 0 and concave down when x > 0Determine the concavity of f(x) = [1/12] x

^{4}− [1/6] x^{3}− x^{2}+ 17x + 23- f′(x) = [1/3] x
^{3}− [1/2] x^{2}− 2x + 17 - f"(x) = x
^{2}− x − 2 - x
^{2}− x − 2 = 0 - (x − 2)(x + 1) = 0
- x = −1 and x = 2 are the points of inflection
- Test points
- f"(−2) = (−2)
^{2}− (−2) − 2 = 4 - f"(0) = (0)
^{2}− 0 − 2 = −2 - f"(3) = 3
^{2}− 3 − 2 = 4

f(x) = [1/12] x

^{4}− [1/6] x^{3}− x^{2}+ 17x + 23 is concave up when x < −1, concave down when −1 < x < 2, and concave up when x > 2*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### First Derivative Test, Second Derivative Test

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Local Minimum and Local Maximum 0:14
- Example
- First and Second Derivative Test 1:26
- First Derivative Test
- Example
- Second Derivative Test (Concavity)
- Example: Concave Down
- Example: Concave Up
- Inconclusive
- Lecture Example 1 5:23
- Lecture Example 2 12:03
- Lecture Example 3 15:54
- Additional Example 4
- Additional Example 5

1 answer

Last reply by: Jamison Czech

Fri Jul 11, 2014 1:51 AM

Post by Jorge Sardinas on March 16, 2014

For the second derivative test wouldn't f''(c) >0 be concave down and f''(c)<0 be concave up, please correct me if I am wrong.

0 answers

Post by Shahaz Shajahan on August 22, 2012

how would you work out the stationary points of sinh(x)?

0 answers

Post by abrar degnh on May 30, 2012

in example two, sorry .

1 answer

Last reply by: alister guerrero

Thu Nov 15, 2012 5:32 PM

Post by abrar degnh on May 30, 2012

i don't think 0 is a critical point because it will be undefined in this example, please correct me if i'm wrong.