Question

Find where it is increasing/decreasing
(5x-4)/(x+4)

Answer 1

I already found that there were no critical points and I'm not sure how to find where it's increasing or decreasing without any critical points

Answer 2

If the graph "rises" as x increases, the function is increasing. Does that help? A function that is increasing certainly can increase without critical points as markers. The CP do define where the function begins to increase (or decrease), however.

Answer 3

Would that mean a positive function is increasing from (-infinity, infinity)?

Answer 4

\(y = \dfrac{5x-4}{x+4} = \dfrac{5(x+4) - 20-4}{x+4} \)

\(= 5 - \dfrac{24}{x+4} \)

\(Y = - \dfrac{24}{x+4} \) where \(Y = y - 5\)

\(Y = - \dfrac{24}{X} \) where \(X = x + 4\)

and so on :)

Answer 5

in terms of critical or fixed points, we have just made linear transformations,

...and find that \(\dfrac{d Y }{dX} = \dfrac{24}{X^2}\)


it follows too that \(\dfrac{d X }{dY} = \dfrac{X^2}{24}\)

Answer 6

Where did you get 5(x+4)-20-4 on the top of the fraction?

Answer 7

\(y = \dfrac{5x-4}{x+4} = \dfrac{\color{red}{5(x+4) - 20-4}}{x+4}\)

\(= \dfrac{\color{red}{5x+20 - 20-4}}{x+4}\)

do you disagree?

Answer 8

I'm just confused why you changed 5x-4 into 5(x+4)-20-4

Answer 9

let me re-think it... if you don't like it

Answer 10

yes, right at the start, i remember.

it's all about simplifying that expression

Answer 11

Your function is y=(5x-4)/(x+4), or \[y=\frac{ 5x-4 }{ x+4 }\]

Answer 12

Jump in and take the derivative immediately:\[\frac{ dy }{ dx }=\frac{ (x+4)(5) - (5x-4)(1) }{ (x+4)^2 }\] Here I used the Quotient Rule. Please be absolutely certain that you understand this; if you don't, ask for clarification.

The derivative simpifies to\[\frac{ dy }{ dx }=\frac{ 5x+20-5x+4 }{ (x+2)^2 }=\frac{ 24 }{ (x+2)^2 }\]

Answer 13

True, you cannot find the critical value(s) by setting this = to 0 and solving for x. HOWEVER, you do have a critical value at x=-2, because the derivative is not defined there.

Plot x=-2 on a number line. On the left you'll have (-inf,-2) and on the right you'll have (-2, inf). From each of these 2 intervals, choose a test number. Subst. this test number into the first derivative to determine whether the deriv. is pos. or neg.

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