Question

A. Find the critical numbers
B. the largest open intervals where the function is increasing
C. the largest open intervals where it is decreasing

y = x (sq root of 9 - x^2)

Answer 1

same problem

Answer 2

\[f(x)=x\sqrt{9-x^2}\] right?

Answer 3

Yeah! I'm having an issue now with after you get from 1/2x (9-x) ^ -1/2

Answer 4

so basically after finding the derivative

Answer 5

whoa what did you get for the derivative?

Answer 6

so I went from
x (9-x) ^1/2

Answer 7

this is a product, you have to use the product rule
\[(fg)'=f'g+g'f\]

Answer 8

ohhhhh

Answer 9

skip the rational exponent, it is not your friend at all in doing this problem

Answer 10

oh okay!

Answer 11

you have
\[(f(x)=x, f'(x)=1,g(x)=\sqrt{9-x^2},g'(x)=?\]
do you know how to find \(g'(x)\) ?

Answer 12

so g'(x) = (9-x^2)^1/2

Answer 13

no

Answer 14

what you did was to rewrite \(g(x)\) with a rational exponent
\[g(x)=\sqrt{9-x^2}=(9-x)^{\frac{1}{2}}\] that is not the derivative
it also does not really help you get it

Answer 15

oh okay. can I use that rewritten form to find the derivative though?

Answer 16

oh wait.. nvm

Answer 17

yes, that and the chain rule
but like i said, i wouldn't recommend it

Answer 18

\[\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\] so by the chain rule
\[\frac{d}{dx}\sqrt{9-x^2}=\frac{1}{2\sqrt{9-x^2}}\times (-2x)=-\frac{x}{\sqrt{9-x^2}}\]

Answer 19

if you want to set something equal to zero and solve you have to use radicals, not negative rational exponents, which will only confuse matters

Answer 20

oh okay. gotcha.

Answer 21

so from there do you use the product rule to find the critical numbers?