anonymous 5 years ago What is the derivative of: (10x-3)/(-2x+4) I get: 34/(-2x+4)^2 is that correct?

1. anonymous

i get 17/2(x-2)^2

2. anonymous

1st step I get (-2x+4)(10) - (10x-3)(-2) / (-2x+4)^2

3. anonymous

you did it right but i factored the bottom out and simplified it

4. anonymous

ok after I get the derivative I need to get the critical number(s). Will that only be a 0

5. anonymous

no its 2 b/c then the denominator = 0

6. anonymous

I see that now Thanks

7. anonymous

You can't have a 0 in the denominator. The function is not differentiable there.

8. anonymous

The critical points will be where the numerator is 0 and the denominator isn't.

9. anonymous

critical numbers the function either equals 0 or is undefined

10. anonymous

so I do not have any critical points then, is that correct?

11. anonymous

Why are you looking for critical points? max/mins? or something different..

12. anonymous

for my problem, I need to: a) find the derivative b) critical number(s) c) Intervals of increase and decrease

13. 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

14. anonymous

To find where the function is increasing/decreasing you just have to see where the derivative is > 0 or < 0 respectivly

15. anonymous

so will I use the 2 on a number line to test to see where it is increasing or decreasing

16. anonymous

yes dude

17. anonymous

or sorry mam

18. anonymous

no problem