Question

What are the critical values of
f(x) = x-1 / x + 3

Answer 1

find the \(\large\color{black}{ f'(x) }\)

Answer 2

ok

Answer 3

is it - x-1 / x + 3

Answer 4

that is not the derivative of your function. or are you trying to say that it id the function?

Answer 5

no I tried to do quotient rule

Answer 6

isnt the quotient rule f'g - fg' / g^2 ??

Answer 7

\(\large\color{black}{ \frac{\LARGE dy}{\LARGE dx} \left(\begin{matrix} \frac{\LARGE x-1}{\LARGE x+3} \\ \end{matrix}\right)=\frac{\LARGE (x+3)(x-1)'-(x+3)'(x-1)}{\LARGE (x+3)^2} }\) makes sense?

Answer 8

so then its (x+3)(1) - (1)(x-1) / (x+3)^2 ?

Answer 9

yes, so the top would be?

Answer 10

wouldny the top be -x-1 because the x+3 divides with one from the denominator

Answer 11

or does it stay 4 / (x+3)^2

Answer 12

yup 4/(x+3)^2

Answer 13

why cant you divide the x+3 if there are two of them

Answer 14

\(\large\color{black}{ f'(x)= \frac{\LARGE 4}{\LARGE (x+3)^2} }\)

Answer 15

when you set the \(\large\color{black}{ f'(x)=0 }\) , is there an x value that satisifes the statement?

Answer 16

please, this isn't your question

Answer 17

@ilovereyvis_x3
\(\large\color{black}{ f'(x)= \frac{\LARGE 4}{\LARGE (x+3)^2} }\)

\(\large\color{black}{ 0= \frac{\LARGE 4}{\LARGE (x+3)^2} }\) .....is there an x-solution?

Answer 18

no

Answer 19

yes, there is no solution.
So there are NO CRITICAL NUMBERS.

Answer 20

\(\large\color{black}{ f'(x)= \frac{\LARGE 4}{\LARGE (x+3)^2} }\) would yield that x=-3 is a critical numb.

but here since -3 is not part of the domain of \(\large\color{black}{ f(x) }\), it is not a critical numb. (x=-3 would have been a critical numb, if it was in the domain of \(\large\color{black}{ f(x) }\) )

Answer 21

ok , but can you help me on how the derivative ended up being 4/ (x+3)^2

Answer 22

\(\large\color{black}{ \frac{\LARGE dy}{\LARGE dx} \left(\begin{matrix} \frac{\LARGE x-1}{\LARGE x+3} \\ \end{matrix}\right)=\frac{\LARGE (x+3)(x-1)'-(x+3)'(x-1)}{\LARGE (x+3)^2} }\) this is the quotent rule.

Answer 23

Correct?

Answer 24

yeah but how is (x+3) - (x-1) = 4???

Answer 25

\(\large\color{black}{ - (x-1) ~~~~~\Rightarrow~~~~~-x+1 }\)

Answer 26

I keep disconnecting, sorry-:(

Answer 27

it's fine don't worry i do also

Answer 28

Again, sjut now

Answer 29

Anyway, see how: \(\large\color{red}{ - (x-1) ~~~~=~~~~-x+1 }\) ?

Answer 30

yes

Answer 31

so,
\(\large\color{red}{ (x+3) - (x-1) = 4 }\)
\(\large\color{red}{ ^{\Huge \color{white}{|}} x+3 - x+1 = 4 }\)

Answer 32

the \(\large\color{red}{X}\)s cancel, and you get 4.

Answer 33

OH !! ok I understand thank you !!!!

Answer 34

anytime:)