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anonymous

  • one year ago

If the tangent line to y = f(x) at (4, 2) passes through the point (0, 1), find f(4) and f '(4). f(4)=2 , f'(4)=?

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  1. myininaya
    • one year ago
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    so you can find the tangent line you are given two points and the slope use point-slope form of a line

  2. anonymous
    • one year ago
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    Y1-Y0=m(x1-xo)?

  3. myininaya
    • one year ago
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    actually you only need the slope

  4. anonymous
    • one year ago
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    yeah

  5. myininaya
    • one year ago
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    since that is what it asked for

  6. myininaya
    • one year ago
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    you know the slope formula right?

  7. myininaya
    • one year ago
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    \[m=\frac{y_1-y_2}{x_1-x_2}\]

  8. anonymous
    • one year ago
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    m=y1-y0/x1-xo

  9. anonymous
    • one year ago
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    is the other way around of how you put it

  10. myininaya
    • one year ago
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    1 and 0's 1 and 2's whatever

  11. myininaya
    • one year ago
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    do you know how to use the formula?

  12. anonymous
    • one year ago
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    is 1/4

  13. myininaya
    • one year ago
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    let's see top is 2-1 and bottom is 4-0 looks great

  14. anonymous
    • one year ago
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    yes

  15. anonymous
    • one year ago
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    do I use y=mx+b?

  16. anonymous
    • one year ago
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    b=2 if I do that with (4,2)

  17. myininaya
    • one year ago
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    you don't need to it just asked to find f'(4) and f(4) and f'(4) is the slope of the tangent line at (4,2) to the curve y=f(x)

  18. myininaya
    • one year ago
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    unless you want to find the tangent line ?

  19. anonymous
    • one year ago
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    I do

  20. myininaya
    • one year ago
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    it is just: \[y-f(a)=f'(a)(x-a) \text{ is the tangent line \to the curve } y=f(x) \text{ at } (a,f(a))\]

  21. myininaya
    • one year ago
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    you are given a is 4 here

  22. myininaya
    • one year ago
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    \[y-f(4)=f'(4)(x-4)\]

  23. myininaya
    • one year ago
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    now you found f(4) and f'(4)

  24. anonymous
    • one year ago
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    right

  25. anonymous
    • one year ago
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    thanks

  26. myininaya
    • one year ago
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    np

  27. anonymous
    • one year ago
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    problem the answer is not 2.

  28. anonymous
    • one year ago
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    I already tried that

  29. myininaya
    • one year ago
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    f(4) should be 2 it is given by (4,2) f'(4) should be 1/4 that is the slope of the tangent line at (4,2) to the curve y=f(x)

  30. anonymous
    • one year ago
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    oh yeah then I was right about the 1/4. in the question I put the f(4)

  31. anonymous
    • one year ago
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    perfect thank you

  32. myininaya
    • one year ago
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    yeah you already found f(4) earlier we were finding f'(4) and we used the slope formula because that is what the derivative means

  33. anonymous
    • one year ago
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    Find f '(a). f(x) = (1 − 2x)^1/2

  34. myininaya
    • one year ago
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    so do you know chain rule?

  35. anonymous
    • one year ago
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    yeah, saw this last year tho

  36. anonymous
    • one year ago
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    I thought it was 1/2(1-2x)^-1/2 * (2)

  37. myininaya
    • one year ago
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    \[h(x)=f(g(x)) \\ h'(x)=g'(x) \cdot f'(g(x)) \\ \\ \text{ if } f(x)=x^n \\ \text{ then we have } \\ h(x)=(g(x))^n \\ h'(x)=g'(x) \cdot n(g(x))^{n-1}\]

  38. myininaya
    • one year ago
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    well one complaint

  39. myininaya
    • one year ago
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    what is the derivative of 1-2x ?

  40. myininaya
    • one year ago
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    you put 2 but it should actually be...

  41. anonymous
    • one year ago
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    ooooh

  42. anonymous
    • one year ago
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    thanks

  43. myininaya
    • one year ago
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    but you can also simplify a bit

  44. anonymous
    • one year ago
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    hahaha hate when that happens

  45. anonymous
    • one year ago
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    yes i did before

  46. anonymous
    • one year ago
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    in my homework

  47. myininaya
    • one year ago
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    ok then you got it now :)

  48. anonymous
    • one year ago
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    do you have time for more?

  49. anonymous
    • one year ago
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    I had problems with this one too

  50. myininaya
    • one year ago
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    maybe one more then I have to go check on something

  51. anonymous
    • one year ago
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    (a) Find the slope m of the tangent to the curve y = 7/ x at the point where x = a > 0.

  52. myininaya
    • one year ago
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    is that just y=7/x?

  53. anonymous
    • one year ago
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    (b) Find equations of the tangent lines at the points (1, 7) and (4, 7/2) .

  54. myininaya
    • one year ago
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    \[y=\frac{7}{x}=7 \cdot \frac{1}{x} =7 x^{-1} \text{ by law of exponents } \\ \text{ now differentiate } y=7 x^{-1} \text{ using power rule }\]

  55. anonymous
    • one year ago
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    no x^1/2 sorry

  56. myininaya
    • one year ago
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    oh

  57. anonymous
    • one year ago
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    it didn't appear when i copy pasted it

  58. anonymous
    • one year ago
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    I had -7/2x^3/2

  59. anonymous
    • one year ago
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    but is not that

  60. myininaya
    • one year ago
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    \[\text{ find the slope of the tangent line \to the curve } y=7x^\frac{1}{2} \text{ at } x=a>0\]?

  61. myininaya
    • one year ago
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    or is that x^(1/2) in the denominator?

  62. anonymous
    • one year ago
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    ok part A i got it

  63. anonymous
    • one year ago
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    I had to change x=a

  64. anonymous
    • one year ago
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    now for B I don't now what I did wrong

  65. myininaya
    • one year ago
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    ok so it was: \[y=\frac{7}{x^\frac{1}{2}}=7x^\frac{-1}{2}\]

  66. anonymous
    • one year ago
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    I got that

  67. anonymous
    • one year ago
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    but I have part B wrong

  68. myininaya
    • one year ago
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    ok the tangent line at (1,7) is \[y-f(a)=f'(a)(x-a) \\ \text{ here you have } a=1 \\ y-f(1)=f'(1)(x-1) \\ \text{ you already found } f'(x) \text{ \above } \\ \text{ you got that } f'(x)=\frac{-7}{2x^{\frac{3}{2}}} \text{ I think but I don't really know anymore } \\ \text{ \because I'm still trying \to figure out what } f(x) \text{ was :p } \\ \text{ so this becomes } \\ y-7=\frac{-7}{2(1)^\frac{3}{2}}(x-1)\]

  69. myininaya
    • one year ago
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    and you still have to do this same process for the other point

  70. anonymous
    • one year ago
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    perfect

  71. anonymous
    • one year ago
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    is not that apparently

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