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INSTRUCTORS Raffi Hovasapian John Zhu
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For more information, please see full course syllabus of Calculus AB
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Lecture Comments (9)

0 answers

Post by Dev Ghan on November 21, 2015

The video keeps cutting off at 11:30. Can anyone fix this? Thanks

0 answers

Post by hamed alsubhi on March 6, 2015

I got the answer for both values positive in example three. how come you got -23.14?

1 answer

Last reply by: Jimmy Wu
Sat Oct 19, 2013 10:56 PM

Post by Si Jia Wen on August 27, 2013

I can't view the end of the video. It keeps cutting me off.

Can someone tell me the answer to example 4? Is it 0 and 3/2?

1 answer

Last reply by: Ben Schalck
Thu Dec 20, 2012 5:38 PM

Post by abbas esmailzadeh on November 3, 2012

is it e^pi/4 and cos^pi/4 or cos of pi/4

0 answers

Post by help me on October 25, 2012

I love your videos :) So easy to understand, unlike Jishi.

1 answer

Last reply by: Arshin Jain
Sat Jan 4, 2014 6:23 PM

Post by Steve Denton on October 12, 2012

on example 3 at 10:28, my graph does not show a minimum at pi, rather at 3.9 or so. The slope is not 0 at pi. Why were not the tests done with the derivative function?

Minimum & Maximum

  • Critical value (maximum or minimum) exists when derivative function = 0
  • Finding critical value
    • Take derivative of appropriate function
    • Set derivative function = 0
    • Solve for x
    • Examine derivative behavior around critical value to determine maximum or minimum
  • Make sure to distinguish between relative and absolute max and min

Minimum & Maximum

Find the minimum or maximum of the following function f(x) = 2x2 − 8x + 8
  • f′(x) = 4x − 8
  • Find where the derivative is zero
  • 4x − 8 = 0
  • 4x = 8
  • x = 2
  • Is this a minimum or a maximum? Test nearby points
  • f′(0) = 4(1) − 8 = −4
  • f′(1) = 4(3) − 8 = 4
  • The derivative goes from negative to positive, so it's a minimum
Minimum at x = 2
Find the relative minimum or maximum of the following function f(x) = −x3 + x2 + x − 1
  • f′(x) = −3x2 + 2x + 1
  • Use quadratic equation to find zeros or roots
  • x = [(−2 ±√{22 − 4(−3)(1)})/(2(−3))]
  • x = [(−2 ±√{16})/(−6)]
  • x = [(−2 ±4)/(−6)]
  • x = −[1/3], 1
  • f′(−1) = −3(−1)2 + 2(−1) + 1 = −4
  • f′(0) = −3(0)2 + 2(0) + 1 = 1
  • f′(2) = −3(2)2 + 2(2) + 1 = −7
Relative minimum at x = −[1/3] and relative maximum at x = 1
Find the minimums and maximums of f(x) = sinx on the interval [0, 2 π]
  • f′(x) = cosx
  • When does cosx equal zero in that interval?
  • f′([(π)/2]) = cos[(π)/2] = 0
  • f′([(3 π)/2]) = cos[(3 π)/2] = 0
  • Test points
  • f′([(π)/4]) = [(√2)/2]
  • f′(π) = −1
  • f′(2 π) = 1
Maximum at x = [(π)/2], minimum at x = [(3 π)/2]
Find the critical points of f(x) = [4/5]x5 − [4/3]x3 + x
  • f′(x) = 4x4 − 4x2 + 1
  • f′(x) = (2x2 − 1)2
  • (2x2 − 1)2 = 0
  • 2x2 − 1 = 0
  • 2x2 = 1
  • x = ±[1/(√2)]
Critical points are x = −[1/(√2)] and x = [1/(√2)]
Find the critical points of f(x) = lnx − x
  • f′(x) = [1/x] − 1
  • f′(1) = [1/1] − 1 = 0
  • We know there's a relative minimum or maximum at x = 1, but there's a variable in the denominator, so it's possible there's an undefined value.
  • f′(0) = [1/0] − 1 is undefined.
Critical points are x = 1 and x = 0
A ball is thrown from the ground into the air. Its position is given by x(t) = −9.8t2 + 19.6t. Find the maximum height the ball reaches.
  • x′(t) = −19.6t + 19.6
  • −19.6t + 19.6 = 0
  • 19.6t = 19.6
  • t = 1
  • It reaches a maximum at t = 1. What position is the ball at at t = 1?
  • x(1) = −9.8(1)2 + 19.6(1) = 9.8
The ball reaches a height of 9.8
Find the maximum and minimum of the following function f(x) = √{1 − x2} in the interval (−1,1)
  • f′(x) = [1/2] [1/(√{1 − x2})] [d/dx] (1 − x2)
  • f′(x) = [(−x)/(√{1 − x2})]
  • f′(0) = [(−0)/(√{1 − 0})] = 0
  • f′(−.5) = [.5/(√{1 − (−.5)2})] > 0
  • f′(.5) = [(−.5)/(√{1 − (.5)2})] < 0
Maximum at x = 0
Find the critical points of the following function f(x) = 2x3 − 6x2 + 6x + 9
  • f′(x) = 6x2 − 12x + 6
  • f′(x) = 6(x2 − 2x + 1)
  • f′(x) = 6(x − 1)2
  • 6(x − 1)2 = 0
  • x − 1 = 0
  • x = 1
Critical point at x = 1
Find the critical points of the following function f(x) = 2x3 − 6x2 + 3x + 99
  • f′(x) = 6x2 − 12x + 3
  • f′(x) = 3(2x2 − 4x + 1)
  • 3(2x2 − 4x + 1) = 0
  • (2x2 − 4x + 1) = 0
  • Quadratic equation
  • x = [(−(−4) ±√{(−4)2 − 4(2)(1)})/2(2)]
  • x = [(4 ±√{16 − 8})/4]
  • x = [(4 ±2√2)/4]
  • x = 1 ±[(√2)/2]
The critical points are x = 1 + [(√2)/2] and x = 1 − [(√2)/2]
Find the relative minimum and maximum of f(x) = x3 − 12x
  • f′(x) = 3x2 − 12
  • 3x2 − 12x = 0
  • x2 − 4x = 0
  • (x + 2)(x − 2) = 0
  • x = −2, 2
  • Test points
  • f′(−3) = 3(−3)2 − 12 = 15
  • f′(0) = 3(0)2 − 12 = −12
  • f′(3) = 3(3)2 − 12 = 15
Relative maximum at x = −2 relative minimum at 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

Minimum & Maximum

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
  • Extrema: First Derivative Test 0:09
    • Example 1
  • Example 2: Real World Application/ Cost Function 4:05
  • Example 3: Minimums & Maximums 7:10
  • Example 4: Find Critical Points 10:52