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Pauloading

  • 3 years ago

How can I linearize that function? f(x) = (x^3 - 2x + 4)^2 where a=1 (step by step)

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  1. Skaematik
    • 3 years ago
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    do you know how to expand a function such as \[\left( x ^{2} +3\right)^{2}\] You multiply the first term by the first term, then 1st by 2nd, then 2nd by 1st, then 2nd by 2nd Do you know what im talking about

  2. hba
    • 3 years ago
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    @Skaematik Can't we use a formula there ?

  3. Skaematik
    • 3 years ago
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    What formula

  4. Pauloading
    • 3 years ago
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    L(x) = f(a) + f'(a)(x-a)

  5. hba
    • 3 years ago
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    Formula to expand (x^2+3)^2 (a+b)^2=a^2+2ab+b^2

  6. hba
    • 3 years ago
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    f(a)= (a^3 - 2a + 4)^2 then, f'(a)= ? Diffrentiate f(a) and find f'(a)

  7. Skaematik
    • 3 years ago
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    Why cant you just expand and take the natural log of both sides?

  8. hba
    • 3 years ago
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    I am not sure about your input @Skaematik

  9. hba
    • 3 years ago
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    @Pauloading Did you differentiate it ?

  10. Pauloading
    • 3 years ago
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    I imagine it's: f'(a) = 2.(3x^2 - 2.1 + 0), right?

  11. hba
    • 3 years ago
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    I imagine that its f'(a) not f'(x)

  12. Skaematik
    • 3 years ago
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    This is an explanation I copied from a site Starting with y = ax^b, we can take logs to any base c on both sides: logc(y) = b logc(x) + logx (a) So the graph of logc y vs. logcx is linear, with slope b and y-intercept equal to logc a.

  13. hba
    • 3 years ago
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    @Pauloading do you follow @Skaematik Method ?

  14. Pauloading
    • 3 years ago
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    Not yet :x

  15. hba
    • 3 years ago
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    @Pauloading Do you want to follow my method ?

  16. Pauloading
    • 3 years ago
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    I'ts sound like greek to me :'( I'm really bad on it.

  17. g152xx
    • 3 years ago
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    if it is f-1(a) when a = 2... set the first equation equal to 2, which will happen when x = 1. So, f-1(2) = 1. Now if you want f-1'(x), you have: = 1 / f'(f-1(x)) so 1/ f'(1) find the derivative of f(X): 3x^2 + 2 so, answer = 1/(3(1) +2) = 1/5. Now just replace a = 1 instead of a = 2. Easy.

  18. hba
    • 3 years ago
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    @g152xx I hope you can help @Pauloading ,i gotta go workout ?

  19. Pauloading
    • 3 years ago
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    Imagine that I've a new function: f(x) = x^3 a=1 and also I try to apply that formula --> L(x) = f(a) + f'(a)(x-a) L(x) = f(a^3) + f'(3a^2)(a^3-1) L(x) = (1^3) + f'(3.1^2)(1^3-1) L(x) = 1 + 3.(0) L(x) = 1 Is it right?

  20. g152xx
    • 3 years ago
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    @hba I'll surely try!

  21. Pauloading
    • 3 years ago
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    Thanks @hba :)

  22. hba
    • 3 years ago
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    Do you know that a linear function has a point (x1,y1) (Let) and a slope m (Let),Anyways Thanks a lot @g152xx And @Pauloading Your'e welcome.

  23. g152xx
    • 3 years ago
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    @Pauloading It sounds right 2me. Unless you wanted the answer in the form L(x) = f(a)+f'(a)(x-a) instead of simplifying it.....

  24. g152xx
    • 3 years ago
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    @Pauloading Bump the question! Let's help him ppl! C'mon!

  25. hartnn
    • 3 years ago
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    we have the formula, L(x) = f(a) + f'(a)(x-a) why don't we use it directly ?? a=1 f(1) = 9 f'(1)=6 L(x) = 9+6(x-1) = 6x +3

  26. Pauloading
    • 3 years ago
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    How did u get 6 at f'?

  27. hartnn
    • 3 years ago
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    could you find f'(x) =... ?

  28. hba
    • 3 years ago
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    I asked her the same thing before :D

  29. Pauloading
    • 3 years ago
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    I got 2.(3x^2 - 2)

  30. hartnn
    • 3 years ago
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    no, thats not correct... u need chain rule.

  31. Pauloading
    • 3 years ago
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    hm....got it

  32. hartnn
    • 3 years ago
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    ok, what u got f'(x) =... ?

  33. hba
    • 3 years ago
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    The answer would be 2 (-2 + 3 a^2) (4 - 2 a + a^3)

  34. hartnn
    • 3 years ago
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    let me verify it for you....

  35. hartnn
    • 3 years ago
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    Imagine that I've a new function: f(x) = x^3 a=1 and also I try to apply that formula --> L(x) = f(a) + f'(a)(x-a) L(x) = f(a^3) + f'(3a^2)(a^3-1) <-----INCORRECT L(x) = (1^3) + f'(3.1^2)(1^3-1) L(x) = 1 + 3.(0) L(x) = 1 L(x) = (1^3) + (3.1^2)(x-1) =3+3(x-1) =3x

  36. Pauloading
    • 3 years ago
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    Now I understood it '-' Thanks everyone :D

  37. hba
    • 3 years ago
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    d/da((a^3-2 a+4)^2) Using the chain rule, d/da((a^3-2 a+4)^2) = 2 u ( du)/( da), where u = a^3-2 a+4 and ( du^2)/( du) = 2 u: 2 (a^3-2 a+4) (d/da(a^3-2 a+4)) Differentiate the sum term by term and factor out constants: 2 (a^3-2 a+4) (d/da(a^3)-2 (d/da(a))+d/da(4)) The derivative of 4 is zero: 2 (a^3-2 a+4) (d/da(a^3)-2 (d/da(a))+0) The derivative of a is 1: 2 (a^3-2 a+4) (d/da(a^3)-2 1) The derivative of a^3 is 3 a^2: Answer: | | 2 (a^3-2 a+4) (3 a^2-2)

  38. hartnn
    • 3 years ago
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    was that ^ copied ?

  39. hartnn
    • 3 years ago
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    from wolf ?

  40. hba
    • 3 years ago
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    Yeah :D

  41. hartnn
    • 3 years ago
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    gotcha.....

  42. hba
    • 3 years ago
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    I had to show her how to do it :)

  43. hba
    • 3 years ago
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    him*

  44. hartnn
    • 3 years ago
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    okay. welcome paul ^_^

  45. hba
    • 3 years ago
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    Paulo*

  46. Pauloading
    • 3 years ago
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    Thanks >.<

  47. hba
    • 3 years ago
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    Your'e Welcome.

  48. hartnn
    • 3 years ago
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    ok,\[ \begin{array}l\color{red}{\text{P}}\color{orange}{\text{a}}\color{#e6e600}{\text{u}}\color{green}{\text{l}}\color{blue}{\text{o}}\color{purple}{\text{ }}\color{purple}{\text{C}}\color{red}{\text{e}}\color{orange}{\text{c}}\color{#e6e600}{\text{í}}\color{green}{\text{l}}\color{blue}{\text{i}}\color{purple}{\text{o}}\color{purple}{\text{}}\end{array} \]

  49. Pauloading
    • 3 years ago
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    LoL :D

  50. hba
    • 3 years ago
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    \[\Huge{\bf{\color{blue}{O}\color{red}{K}}}\]

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