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onegirl Group Title

Write the equation (estimated) of the tangent line.h(x) = f(x)g(x) at x = 1.

  • one year ago
  • one year ago

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  1. onegirl Group Title
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    @rizwan_uet can u help?

    • one year ago
  2. Mertsj Group Title
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    Would you please post the problem correctly?

    • one year ago
  3. onegirl Group Title
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    okay

    • one year ago
  4. onegirl Group Title
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    \[h(x) = f(x)g(x) \] at x = 1

    • one year ago
  5. onegirl Group Title
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    @tcarroll010 can u help?

    • one year ago
  6. onegirl Group Title
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    @RadEn can u help?

    • one year ago
  7. Mertsj Group Title
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    How is anyone supposed to help you when we have no idea what the functions are?

    • one year ago
  8. onegirl Group Title
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    well thats the question, it says find an equation of a tangent line to h(x) = f(x)g(x) at x = 1

    • one year ago
  9. onegirl Group Title
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    there are no functions..

    • one year ago
  10. onegirl Group Title
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    so can anyone help?

    • one year ago
  11. tcarroll010 Group Title
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    You can get the first derivative of h(x) as follows with the product rule: h'(x) = f'(x)g(x) + f(x)g'(x) That will be your slope at any point where the first derivative is defined. Now, you use the point-slope formula for the equation of a line:\[y - y _{1} = m(x - x _{1})\]and here, "m" is as defined above with the first derivative for your slope. Make sure you substitute "1" where you see "x" in that first equation. x1 is 1 and y1 is h(1). So, just make the substitutions and that is the equation for the tangent line.

    • one year ago
  12. onegirl Group Title
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    I did this but my teacher told me to write the equation an estimated one "the curve is (1, h(1)) , using the product rule h'(x) = f'(x) g'(x) = + g(x)f'(x) substituting x= 1 into the 1st derivative the slope will be h'(1) = f(1) g'(1) + g(1)f'(1) so the slope h'(1) and the point (1,h(1) so y = (f'(1)g(1) +f(1)g'(1) (x - 1) + f(1)g(1)

    • one year ago
  13. onegirl Group Title
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    so first i have to get the derivative of h'(x) f'(x) g(x) + f(x)g'(x) right?

    • one year ago
  14. tcarroll010 Group Title
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    his is just what I said also. Exactly what I said.

    • one year ago
  15. onegirl Group Title
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    okay

    • one year ago
  16. tcarroll010 Group Title
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    The only difference, which is really nothing, is that I said y1 is h(1). That is the same as f(1)g(1). Just written slightly differently, that's all. And that's all you have to do. And you put the h(1) on the right which is fine. Same answer.

    • one year ago
  17. onegirl Group Title
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    Well i did this but my teacher said i need to write an equation and estimated one. :/

    • one year ago
  18. onegirl Group Title
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    so maybe my final answer needs to be estimated?

    • one year ago
  19. tcarroll010 Group Title
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    Yours and my answer, which is the same and is the correct and exact answer. Let me see if we can estimate this for x = 1. Hold on a minute.

    • one year ago
  20. onegirl Group Title
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    okay

    • one year ago
  21. tcarroll010 Group Title
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    This can be re-written: y = [f'(1)g(1) + f(1)g'(1)](x - 1) + f(1)g(1) y = [f'(1)g(1) + f(1)g'(1)]x + ( f(1)g(1) - [f'(1)g(1) + f(1)g'(1)] ) Now, if the derivative at x = 1 is small, we could drop it out of the far right side of the above equation as it is in the form: y = mx + b y = [f'(1)g(1) + f(1)g'(1)]x + f(1)g(1) But that is if the derivative at x = 1 is small. Then, we have a good approximation.

    • one year ago
  22. onegirl Group Title
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    okay

    • one year ago
  23. onegirl Group Title
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    well i'll try this thanks

    • one year ago
  24. tcarroll010 Group Title
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    If I were the teacher, this is what I'd be looking for. Remember the stipulation of the derivative being small.

    • one year ago
  25. onegirl Group Title
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    okay

    • one year ago
  26. tcarroll010 Group Title
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    Good luck to you in all of your studies and thx for the recognition! @onegirl And you're welcome!

    • one year ago
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