Question

identify extrema f(x) = xsqrt {4-x^2}

Answer 1

Take a derivative and set it to zero. You must do a product rule and a chain rule.
f'(x)=(1)sqrt(4-x^2)+(x)(1/2)(-2x)(4-x^2)^(-1/2)
Set that to zero.

Answer 2

ok i cant really make sense of what you typed but ig got 4- 2x^2 / sqrt{4-x^2}

Answer 3

ok i got the same thing when i multiply it out but my problem is when i identify intervals of increase and decrease if its a square root and the interval im using is negative for my sign chart do i say DNE or negative?

Answer 4

Let me type it in latex. H/o\[f'(x)=(1)\sqrt{4-x^2}+\frac{(x)(\frac{1}{2})(-2x)}{\sqrt{4-x^2}}\].

Answer 5

like ot (-\[\infty\], -\[\sqrt{2}\]) and when i plug in -3 for \[\sqrt{4-x ^{2}}\] i get a negative so whats my sign? negative right?

Answer 6

thats my interval sorry \[(-\infty, -\sqrt{2})\]

Answer 7

Well, find the zeros and D.N.E.'s first. Then from there, test to see if they are max/mins. Once you have that. You can find the intervals of decreasing/increasing. I'll tell you what I get. 2 seconds.

Answer 8

kk

Answer 9

Okay, I got x=sqrt(2) and -sqrt(2). So I would plug in -2 and -1 to test -sqrt(2) and 1 and 2 to test the other.
Increasing to increasing and decreasing to decreasing means no max/min. Increasing to decreasing, or decreasing to increasing does. From there, you can max your sign chart (don't use 3, thats outside of the domain (sqrt(4-(3)^2) is complex)

Answer 10

I suppose 2 technically is too :/

Answer 11

i used -3

Answer 12

But you can't. Because you get a negative in the square root. And we aren't considering complex differentiation.

Answer 13

i mean like when i used the sign chart for the interval \[(-\infty, -\sqrt{2})\]
yeah so what can I do there?

Answer 14

But you don't have that interval, because (-infty,-2) is outside of the domain same with (2,infty)

Answer 15

http://www.wolframalpha.com/input/?i=extrema+of+y%3Dxsqrt(4-x^2) Look at the graph. You'll see what I mean.

Answer 16

so i used those intervals

Answer 17

ok i see...thank you

Answer 18

No problem. Just use the interval (-2,-sqrt(2)) then (-sqrt(2),sqrt(2)) then (sqrt(2),2)

Answer 19

intervals**

Answer 20

okay thank you so much

Answer 21

No problem :P