Question

What is the derivative of:
(10x-3)/(-2x+4)

I get: 34/(-2x+4)^2

is that correct?

Answer 1

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

Answer 2

1st step I get

(-2x+4)(10) - (10x-3)(-2) / (-2x+4)^2

Answer 3

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

Answer 4

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

Answer 5

no its 2 b/c then the denominator = 0

Answer 6

I see that now
Thanks

Answer 7

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

Answer 8

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

Answer 9

critical numbers the function either equals 0 or is undefined

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

\[\frac{d}{dx}(10x-3)(-2x+4)^{-1}\]
\[=(10x-3)\frac{d}{dx}[(-2x + 4)^{-1}] + (-2x+4)^{-1}\frac{d}{dx}[10x-3]\]
\[=(10x-3)[2(-2x+4)^{-2}] + 10(-2x+4)^{-1}\]
\[=\frac{2(10x-3)}{(-2x+4)^2} + \frac{10(-2x+4)}{(-2x+4)^2}\]
\[\text{ Critical points are solutions to: }\]
\[2(10x-3) + 10(-2x+4) = 0 \text{, } x \ne 2 \]

\[\rightarrow 20x -6 -20x + 40 = 0 \rightarrow \text{No solutions}\]

Function has a vertical asymptote at x=2

Answer 14

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

Answer 15

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

Answer 16

yes dude

Answer 17

or sorry mam

Answer 18

no problem