Calculus 3.10 Linear Approximations and Differentials

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  • เผยแพร่เมื่อ 2 ม.ค. 2025

ความคิดเห็น • 8

  • @potatopowers4602
    @potatopowers4602 ปีที่แล้ว +1

    00:00 Linear Approximation
    01:57 Example 1
    06:58 Example 2
    13:25 Independent Variables
    15:25 Example 3
    19:04 Example 4

  • @SumitKothari700
    @SumitKothari700 ปีที่แล้ว +1

    thank you for this playlist man

  • @willardsavage2980
    @willardsavage2980 ปีที่แล้ว

    excellent videos im leaving my like in each one of them

  • @LolBruh
    @LolBruh 2 ปีที่แล้ว +4

    How did 3.98 become .98 and how did 4.05 become 1.05 when you plugged them in? So confused

    • @asherroberts
      @asherroberts  2 ปีที่แล้ว +4

      I used √x+3 to approximate √3.98 and √4.05. Therefore, I have to plug in the values of x that will make √x+3 equal to √3.98 and √4.05. Since 0.98+3=3.98 and 1.05+3=4.05, I use 0.98 and 1.05 for x.

    • @LolBruh
      @LolBruh 2 ปีที่แล้ว

      @@asherroberts Wow! That makes a lot of sense; thank you!

  • @yA2327R
    @yA2327R 4 ปีที่แล้ว +3

    is it possible to do what you did on the calculator without the calculator?

    • @asherroberts
      @asherroberts  4 ปีที่แล้ว +3

      Yes, you could carefully graph the functions by hand instead of using a calculator, use a table of values, or solve the equation 7/4+x/4=sqrt(x+3)+0.5 algebraically, but these are all fairly cumbersome methods. The question is designed for use with a calculator.