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joycelefebvre Group Title

find slope of tangent line y=f(x)=xsquared + 1 at (2,5) given f'(x)=2x

  • 2 years ago
  • 2 years ago

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  1. cwrw238 Group Title
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    the slope = 2x at (2,5) x = 2 so slope of tangent = 2(2) = 4

    • 2 years ago
  2. 2bornot2b Group Title
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    So your question is to find the slope of the equation\(f(x)=x^2+1\) at (2,5) given f'(x)=2x

    • 2 years ago
  3. 2bornot2b Group Title
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    joycelefebvre are you new to OS?

    • 2 years ago
  4. joycelefebvre Group Title
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    yes could you show me your work on that please i'm studying for praxis 0061 my study guied doesnt show work its horrible tu

    • 2 years ago
  5. 2bornot2b Group Title
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    OK let me help you

    • 2 years ago
  6. 2bornot2b Group Title
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    Given any function, the derivative of the function is the equation to the slope of the given equation--------did you know that?

    • 2 years ago
  7. joycelefebvre Group Title
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    kinda I took calc in college in 93 and its not coming back

    • 2 years ago
  8. 2bornot2b Group Title
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    OK, then I am trying to refresh that. Suppose you have a function y=f(x), so the function y=f'(x) is the equation of the tangent of the function y=f(x)

    • 2 years ago
  9. 2bornot2b Group Title
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    OK?

    • 2 years ago
  10. 2bornot2b Group Title
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    So your given function is \(f(x)=x^2+1\) and \(f'(x)=2x\)

    • 2 years ago
  11. 2bornot2b Group Title
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    Now as I told you earlier, \(f'(x)\) should be the equation of the tangent at any point

    • 2 years ago
  12. 2bornot2b Group Title
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    Your question asks you to find the equation of the tangent at (2,5) so just put x=2 in the equation to tangent i.e. 2x2=4. So the answer is 4 Got it?

    • 2 years ago
  13. joycelefebvre Group Title
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    omg I was trying to do y2-y1/x2-x1

    • 2 years ago
  14. 2bornot2b Group Title
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    This is actually a theorem.. If you just remember it you don't have to find the tangent that way.......... I mean the way you are trying using the formula \(\frac {y2-y1}{x2-x1}\)

    • 2 years ago
  15. 2bornot2b Group Title
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    The theorem is, Let y=f(x) be a given equation, to find the equation to the tangent at any point is y=f'(x) where f'(x) is the derivative of the function f(x)

    • 2 years ago
  16. 2bornot2b Group Title
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    Is that clear, any problem?

    • 2 years ago
  17. joycelefebvre Group Title
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    tu i couldn't find the praxis study guied for 0061 so i got the cliff notes and its all text no examples

    • 2 years ago
  18. 2bornot2b Group Title
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    Did you know you can chat with other users in the maths group?

    • 2 years ago
  19. 2bornot2b Group Title
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    In that way you can get help from experts?

    • 2 years ago
  20. 2bornot2b Group Title
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    Come join me in the chat room (do you know how?)

    • 2 years ago
  21. joycelefebvre Group Title
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    no

    • 2 years ago
  22. 2bornot2b Group Title
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    Do you see those green boxes below

    • 2 years ago
  23. joycelefebvre Group Title
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    your 31

    • 2 years ago
  24. 2bornot2b Group Title
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    See there I have posted you a chat message

    • 2 years ago
  25. 2bornot2b Group Title
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    Yes, I am 31, thats my level.

    • 2 years ago
  26. 2bornot2b Group Title
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    But chatting is a different thing

    • 2 years ago
  27. joycelefebvre Group Title
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    my lap top keeps telling me i cant run this script

    • 2 years ago
  28. 2bornot2b Group Title
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    OH, I see, what operating system are you using?

    • 2 years ago
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