Question

Critical points.

Answer 1

why u all hate math today? lol

Answer 2

I do know what they are... just confirming my solution

Answer 3

\[f(w) = {x^2 + 1\over x^2 - x - 6} \]I should just equate both the numerator and denominator to 0 for getting the critical points, right?

Answer 4

\(f(x)\)**

Answer 5

I always hated Calculus -.-

Answer 6

In calculus, a critical point of a function of a real variable is any value in the domain where either the function is not differentiable or its derivative is 0

Answer 7

so first derivate f(x) then equate it to 0.

Answer 8

and ur not allowed to division by zero...so denum must be nonzero

Answer 9

@hartnn Thank you, but can you please look for my solution above and check it?

Answer 10

Yes, yes I stated that in my solution as well!

Answer 11

there is no solution above.
I should just equate both the numerator and denominator to 0 for getting the critical points, <----incorrect.

Answer 12

Please look at my solution and check it.

Answer 13

Ok.

Answer 14

have u learned u/v rule in derivative ??

Answer 15

Should I first derive the function?

Answer 16

Yes, quotient rule.

Answer 17

then apply that first.

Answer 18

\[\left(x^2+ 1\over x^2 - x + 6 \right)' = {2x(x^2 - x+ 6) - (x^2 + 1)} \]

Answer 19

I got the notation wrong... but still

Answer 20

if u want to check your answer,check here for derivative.
and yes,critical points are minimas and maximas,mentioned here,check those are your final critical points.
http://www.wolframalpha.com/input/?i=%7Bx%5E2+%2B+1%5Cover+x%5E2+-+x+-+6%7D

Answer 21

\[ f(x)=\left(x^2+ 1\over x^2 - x + 6 \right)' \]\[f'(x) = {2x(x^2 - x+ 6) - (x^2 + 1)}\]

Answer 22

\[\implies 2x^3 - 2x^2 + 12x - x^2 - 1 \]\[\implies 2x^3 - 3x^2 + 12x - 1 \]

Answer 23

Oh wait

Answer 24

Never mind, I got it!

Answer 25

That solution which I showed was wrong... but I can manage the rest. Thank you!!