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NathanJHW

  • one year ago

The function f (x) has the value f (−1) = 1. The slope of the curve y = f (x) at any point is given by the expression dy/dx = (2x + 1)( y + 1)

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  1. NathanJHW
    • one year ago
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    A. Write an equation for the line tangent to the curve y = f (x) at x = − 1. B. Use the tangent line from part A to estimate f (−0.9) C. Use separation of variables to find an explicit or implicit formula for y = f (x), with no integrals remaining. D. Find lim f (x) as x approaches infinity.

  2. NathanJHW
    • one year ago
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    @jim_thompson5910

  3. jim_thompson5910
    • one year ago
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    we're told that f(-1) = 1 so that means (-1,1) lies on the graph of f(x)

  4. jim_thompson5910
    • one year ago
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    plug in x = -1 and y = 1 into the dy/dx formula and that will determine the slope m

  5. NathanJHW
    • one year ago
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    -1?

  6. NathanJHW
    • one year ago
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    -2 sorry

  7. jim_thompson5910
    • one year ago
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    nope on either one

  8. NathanJHW
    • one year ago
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    How do I do it then?

  9. jim_thompson5910
    • one year ago
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    you plugged it into dy/dx = (2x + 1)( y + 1) right?

  10. NathanJHW
    • one year ago
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    (2(-1)+1)(1+1)

  11. jim_thompson5910
    • one year ago
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    oh wait, I divided on accident for some reason

  12. jim_thompson5910
    • one year ago
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    idk why I did that

  13. jim_thompson5910
    • one year ago
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    yeah you're right, m = -2

  14. jim_thompson5910
    • one year ago
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    anyways, we know the following x = -1 y = 1 m = -2 now plug those into y = mx+b and solve for b

  15. NathanJHW
    • one year ago
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    b= -1

  16. jim_thompson5910
    • one year ago
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    so the tangent line is y = -2x - 1

  17. NathanJHW
    • one year ago
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    Alright, now how do we do B?

  18. jim_thompson5910
    • one year ago
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    plug in x = -0.9 into the tangent line equation to get the approximate value for f(-0.9)

  19. NathanJHW
    • one year ago
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    -2.8

  20. jim_thompson5910
    • one year ago
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    the idea is that if we plug in values close to x = -1, then we'll get rough approximations to f(x) be warned: the further you stray from x = -1, the larger the error will be until it will be so large that it's not worth it

  21. jim_thompson5910
    • one year ago
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    -2.8? you sure?

  22. NathanJHW
    • one year ago
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    -2(0.9)-1

  23. jim_thompson5910
    • one year ago
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    negative 0.9

  24. NathanJHW
    • one year ago
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    oops

  25. NathanJHW
    • one year ago
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    0.8

  26. jim_thompson5910
    • one year ago
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    that's the approx value of f(-0.9)

  27. NathanJHW
    • one year ago
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    Now how do we do part C?

  28. jim_thompson5910
    • one year ago
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    \[\Large \frac{dy}{dx} = (2x+1)(y+1)\] \[\Large dy = (2x+1)(y+1)dx\] \[\Large \frac{dy}{y+1} = (2x+1)dx\] \[\Large \int\frac{dy}{y+1} = \int(2x+1)dx\] \[\Large \ln(|y+1|) = x^2+x+C\] I'll let you finish up. The goal is to isolate y (which is the same as finding f(x))

  29. NathanJHW
    • one year ago
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    Is it y = e^(C+x^2+x)-1

  30. jim_thompson5910
    • one year ago
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    \[\Large \ln(|y+1|) = x^2+x+C\] \[\Large |y+1| = e^{x^2+x+C}\] \[\Large |y+1| = e^C*e^{x^2+x}\] Let's find the value of e^C using x = -1 and y = 1 \[\Large |y+1| = e^C*e^{x^2+x}\] \[\Large |1+1| = e^C*e^{(-1)^2+(-1)}\] \[\Large |2| = e^C*e^{0}\] \[\Large 2 = e^C*1\] \[\Large 2 = e^C\] \[\Large e^C = 2\] ------------------------------------------------------- So \[\Large |y+1| = e^C*e^{x^2+x}\] turns into \[\Large |y+1| = 2*e^{x^2+x}\]

  31. jim_thompson5910
    • one year ago
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    The absolute value is often a pain, but the good news is that because the right side is never negative, we can drop the absolute values. So solving for y gives \[\Large |y+1| = 2*e^{x^2+x}\] \[\Large y+1 = 2*e^{x^2+x}\] \[\Large y = 2*e^{x^2+x}-1\] \[\Large f(x) = 2*e^{x^2+x}-1\]

  32. NathanJHW
    • one year ago
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    Alright, sorry I had to go somewhere @jim_thompson5910 back now though.

  33. NathanJHW
    • one year ago
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    For the last one, f(x) is the answer to c right? @jim_thompson5910

  34. jim_thompson5910
    • one year ago
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    yeah hopefully you see how it all leads up to that

  35. NathanJHW
    • one year ago
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    I got infinity

  36. jim_thompson5910
    • one year ago
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    me too

  37. NathanJHW
    • one year ago
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    alright thank you so much for all your help @jim_thompson5910 really appreciate it!

  38. jim_thompson5910
    • one year ago
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    np

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