Question

how do i find the critical numbers of x^2/x^2-9

Answer 1

Just checking that the equation is \[\frac{x^2}{x^2-9}\]

If that is the case we want to find the critical values. To find them try to get the bottom of the fraction to equal 0

Answer 2

i hope u'll get ur ans ..f(x) = x/(x^2-9)

Apply the quotient rule:
f'(x) = [(x^2-9)(1)-x (2x)] /(x^2-9)^2
f'(x) = ( x^2-9 -2x^2) /(x^2-9)^2
f'(x) = (-x^2-9) /(x^2-9)^2 = 0

-x^2-9=0
x^2=-9 has no solution.
Because when x=-3 or 3 the denominator is undefined, the critical points are:
x=-3 and x=3

Answer 3

The critical points can not be 3 or -3

Answer 4

when you plug in 3 and -3 into the function you will get 9/0 which is undefined therefore \[x=\pm3\] can NOT be the critical points

Answer 5

the critical point is x=0 when you completely simplify out the derivative you will get -18x/(x^2-9)=0

Answer 6

Well I feel silly. I agree though, good job.

Answer 7

where you first find when the numerator is equal to 0. -18x=0, x=0 therefore x=0 is your only critical number

Answer 8

when an x value gives you an undefined number you can't use it as a critical point

Answer 9

i hope you understand this?!

Answer 10

\[f(x)=\frac{x^2}{x^2-9}\]
\[f'(x)=\frac{2x(x^2-9)-2x^3}{(x^2-9)^2}\]
\[=-\frac{18x}{(x^2-9)^2}\]
\[f''(x)=\frac{-18(x^2-9)^2+18x(2)(2x)(x^2-9)}{(x^2-9)^4}\]
\[f''(x)=\frac{-18(x^2-9)+72x^2}{(x^2-9)^3}\]
\[\large =\frac{54x^2+162}{(x^2-9)^3}\]
At S.P's (Stationary Points) f'(x)=0
\[\large -18x=0\]
\[\large x=0\]
\[\large y=0\]
\[f''(x)<0\]
Therefore Maximum T.P(Turning Point) at (0, 0)
At I.P's (Inflexion Point) f''(x)=0
\[\large 54x^2=-162\]
Therefore no I.P's.