anonymous
  • anonymous
What is the derivative of: (10x-3)/(-2x+4) I get: 34/(-2x+4)^2 is that correct?
Mathematics
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anonymous
  • anonymous
What is the derivative of: (10x-3)/(-2x+4) I get: 34/(-2x+4)^2 is that correct?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
i get 17/2(x-2)^2
anonymous
  • anonymous
1st step I get (-2x+4)(10) - (10x-3)(-2) / (-2x+4)^2
anonymous
  • anonymous
you did it right but i factored the bottom out and simplified it

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anonymous
  • anonymous
ok after I get the derivative I need to get the critical number(s). Will that only be a 0
anonymous
  • anonymous
no its 2 b/c then the denominator = 0
anonymous
  • anonymous
I see that now Thanks
anonymous
  • anonymous
You can't have a 0 in the denominator. The function is not differentiable there.
anonymous
  • anonymous
The critical points will be where the numerator is 0 and the denominator isn't.
anonymous
  • anonymous
critical numbers the function either equals 0 or is undefined
anonymous
  • anonymous
so I do not have any critical points then, is that correct?
anonymous
  • anonymous
Why are you looking for critical points? max/mins? or something different..
anonymous
  • anonymous
for my problem, I need to: a) find the derivative b) critical number(s) c) Intervals of increase and decrease
anonymous
  • anonymous
\[\frac{d}{dx}(10x-3)(-2x+4)^{-1}\] \[=(10x-3)\frac{d}{dx}[(-2x + 4)^{-1}] + (-2x+4)^{-1}\frac{d}{dx}[10x-3]\] \[=(10x-3)[2(-2x+4)^{-2}] + 10(-2x+4)^{-1}\] \[=\frac{2(10x-3)}{(-2x+4)^2} + \frac{10(-2x+4)}{(-2x+4)^2}\] \[\text{ Critical points are solutions to: }\] \[2(10x-3) + 10(-2x+4) = 0 \text{, } x \ne 2 \] \[\rightarrow 20x -6 -20x + 40 = 0 \rightarrow \text{No solutions}\] Function has a vertical asymptote at x=2
anonymous
  • anonymous
To find where the function is increasing/decreasing you just have to see where the derivative is > 0 or < 0 respectivly
anonymous
  • anonymous
so will I use the 2 on a number line to test to see where it is increasing or decreasing
anonymous
  • anonymous
yes dude
anonymous
  • anonymous
or sorry mam
anonymous
  • anonymous
no problem

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