Question

critical point of f(x) -3/x^2+9

Answer 1

Derivative set equal to zero and solve.

Answer 2

The derivative is\[-3 (x ^{2}+9)(2x)\]

Answer 3

x^2+9 is the denominator?

Answer 4

Yes, to the -2, forgot to add that.

Answer 5

yes

Answer 6

use the quotient rule

Answer 7

6x/(x^2+9x)2??

Answer 8

i mean 6x/(x^2+9x)^2?

Answer 9

\[f'(x)=\frac{-3(2x)}{(x^2+9)^2}=\frac{-6x}{(x^2+9)^2}\]

Answer 10

@Dockworker has it

Answer 11

actually, numerator is positive

Answer 12

\[f"(x)=\frac{6x}{(x^2+9)^2}\]

Answer 13

Yes, forgot the negative in front of the 3

Answer 14

f'(x), not f''(x) rather :)

Answer 15

set that equal to zero?

Answer 16

Yes

Answer 17

yes, normally for critical numbers you'd find the zeros and when the derivative is undefined, but because the denominator can never be 0 in this case, there's only one critical number

Answer 18

is it 0?

Answer 19

If you plug it into Wolfram Alpha you can check your answer.

Answer 20

they gave me 0

Answer 21

thank you