Math Questions and Answers

Start Your Free Trial

`y = xsqrt(16-x^2)` Identify the open intervals on which the function is increasing or decreasing.

Expert Answers info

violy eNotes educator | Certified Educator

calendarEducator since 2010

write176 answers

starTop subjects are Math, Science, and Literature

We need to take the derivative of the function first in order to identify the critical points or critical numbers, which we can use for the endpoints of our intervals.

For the right side we need to apply the Product Rule:

` f'(uv) = u'v+v'u ` 

Let us set `u = x`   and  `v= sqrt(16 - x^2)`

Therefore, `u' = 1`   and `v' = 1/2(16-x^2)^(-1/2)*(-2x) = -(x/(16-x^2)^(1/2))`

So, we will have:

`y' = 1*sqrt(16-x^2) +(-x/(16-x^2)^(1/2))*x`  

`y' = sqrt(16-x^2) - x^2/(16-x^2)^(1/2)`

Take note that is the same `(16-x^2)^(1/2)` as `sqrt(16-x^2)` .

We will equate it to zero to find the critical numbers.

`sqrt(16-x^2) - x^2/sqrt(16-x^2) =0`

Multiply both sides by `sqrt(16-x^2)` .

 `sqrt(16-x^2)*sqrt(16-x^2) +sqrt(16-x^2)(-x^2/sqrt(16-x^2))=sqrt(16-x^2)*0`

Remember that `sqrt(a)*sqrt(a) = a` for instance, `sqrt(3)*sqrt(3) = sqrt(3*3) = sqrt(9)=...

(The entire section contains 403 words.)

Unlock This Answer Now



check Approved by eNotes Editorial