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anonymous

  • 5 years ago

What is the derivative of: (10x-3)/(-2x+4) I get: 34/(-2x+4)^2 is that correct?

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  1. anonymous
    • 5 years ago
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    i get 17/2(x-2)^2

  2. anonymous
    • 5 years ago
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    1st step I get (-2x+4)(10) - (10x-3)(-2) / (-2x+4)^2

  3. anonymous
    • 5 years ago
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    you did it right but i factored the bottom out and simplified it

  4. anonymous
    • 5 years ago
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    ok after I get the derivative I need to get the critical number(s). Will that only be a 0

  5. anonymous
    • 5 years ago
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    no its 2 b/c then the denominator = 0

  6. anonymous
    • 5 years ago
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    I see that now Thanks

  7. anonymous
    • 5 years ago
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    You can't have a 0 in the denominator. The function is not differentiable there.

  8. anonymous
    • 5 years ago
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    The critical points will be where the numerator is 0 and the denominator isn't.

  9. anonymous
    • 5 years ago
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    critical numbers the function either equals 0 or is undefined

  10. anonymous
    • 5 years ago
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    so I do not have any critical points then, is that correct?

  11. anonymous
    • 5 years ago
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    Why are you looking for critical points? max/mins? or something different..

  12. anonymous
    • 5 years ago
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    for my problem, I need to: a) find the derivative b) critical number(s) c) Intervals of increase and decrease

  13. anonymous
    • 5 years ago
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    \[\frac{d}{dx}(10x-3)(-2x+4)^{-1}\] \[=(10x-3)\frac{d}{dx}[(-2x + 4)^{-1}] + (-2x+4)^{-1}\frac{d}{dx}[10x-3]\] \[=(10x-3)[2(-2x+4)^{-2}] + 10(-2x+4)^{-1}\] \[=\frac{2(10x-3)}{(-2x+4)^2} + \frac{10(-2x+4)}{(-2x+4)^2}\] \[\text{ Critical points are solutions to: }\] \[2(10x-3) + 10(-2x+4) = 0 \text{, } x \ne 2 \] \[\rightarrow 20x -6 -20x + 40 = 0 \rightarrow \text{No solutions}\] Function has a vertical asymptote at x=2

  14. anonymous
    • 5 years ago
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    To find where the function is increasing/decreasing you just have to see where the derivative is > 0 or < 0 respectivly

  15. anonymous
    • 5 years ago
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    so will I use the 2 on a number line to test to see where it is increasing or decreasing

  16. anonymous
    • 5 years ago
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    yes dude

  17. anonymous
    • 5 years ago
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    or sorry mam

  18. anonymous
    • 5 years ago
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    no problem

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