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anonymous

  • 4 years ago

how do I find the slope of the curve at x = -9. when Let f(x)=5 x^2 + 2/x^2. (this new to me)

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  1. anonymous
    • 4 years ago
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    I first find the f'(x) = 10x+1/-2x right ?

  2. anonymous
    • 4 years ago
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    or dy/dx= 10x-4x^-3 right

  3. anonymous
    • 4 years ago
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    then I plug in for x=-9

  4. dumbcow
    • 4 years ago
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    yes correct

  5. anonymous
    • 4 years ago
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    then it is 10(-9)-4(-9)^3=

  6. anonymous
    • 4 years ago
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    -90-2916=

  7. dumbcow
    • 4 years ago
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    oh the exponent is neg, so you are dividing by 9^3

  8. anonymous
    • 4 years ago
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    wait is ot 10(-9)-4(-9)^-3

  9. dumbcow
    • 4 years ago
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    that is correct --> -90 +4/9^3

  10. dumbcow
    • 4 years ago
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    (-9)^-3 = -1/9^3

  11. anonymous
    • 4 years ago
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    wait im lost

  12. anonymous
    • 4 years ago
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    so its -90-1/9^3

  13. dumbcow
    • 4 years ago
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    you did something wrong when you wrote ...-90 -2916 property of neg exponents: x^-n = 1/x^n right

  14. anonymous
    • 4 years ago
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    yes

  15. dumbcow
    • 4 years ago
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    10(-9) - 4(-9)^-3 = -90 +4/9^3 = -90 +4/729 = -89.9945

  16. anonymous
    • 4 years ago
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    oh ok divide 4/729

  17. anonymous
    • 4 years ago
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    I was multiplying 4(9)^3

  18. dumbcow
    • 4 years ago
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    :)

  19. anonymous
    • 4 years ago
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    is that the approximate or exact answer?

  20. anonymous
    • 4 years ago
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    how can i find the exact number because they want me to find the exact anwser

  21. anonymous
    • 4 years ago
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    nevermind I found it thanx

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