## anonymous 3 years ago help :( locate all critical points ( both types ) of h(x)= radical 8-6*x^2 The critical point(s) is (are) .

1. anonymous

So the function h(x) = $\sqrt{8-6x^2}$ , is that right? By critical points, do you mean relative maxima / minima?

2. anonymous

yes

3. anonymous

i mean x and y

4. anonymous

critical points..

5. anonymous

Oh, okay. First of all, do you know how to take the derivative of h(x)?

6. anonymous

so far im here: -12x/ 2 radical 8-6x^2

7. anonymous

The derivative of h(x) is: $\frac{ -6x }{ \sqrt{8-6x ^{2}} }$

8. anonymous

The next step, is finding where f'(x) = 0 or is undefined, since that's going to find the critical points of f. And you can use those to form intervals to test for minima and maxima.

9. anonymous

So, you're going to set f'(x) to 0 and solve for x.

10. anonymous

okay makes sense... then to find the critical points i equal that to zero... and just solve for x?

11. anonymous

oh so wait, i don't equal it to zero.... i plug in zero into x?? and from the original problem i plug int the x when i find that out?

12. anonymous

$\frac{ -6x }{ \sqrt{8-6x ^{2}} } = 0$ $(\sqrt{8-6x ^{2}}) *\frac{ -6x }{ \sqrt{8-6x ^{2}} } = 0 \sqrt{8-6x ^{2}}$ $-6x = 0$ $x=0$

13. anonymous

Also f is undefined when:$x = \pm \frac{ -2\sqrt{3} }{ 3 }$ and since there can't be a negative square root for the denominator, the domain of f is $[x = \pm \frac{ -2\sqrt{3} }{ 3 }]$ , so you will use that along with 0 to make test intervals.

14. anonymous

I got zero as well... not how do i find y

15. anonymous

i plugged in zero into the normal question... and got 2 radical 2

16. anonymous

17. anonymous

Gah, I was doing something more complicated. Ignore the part about test intervals, sorry. :(

18. anonymous

its okay can we do after we find the x

19. anonymous

the y..

20. anonymous

Plug in the critical numbers into the f(x).

21. anonymous

i plug in 0 in the original function? to find y right

22. anonymous

Yes, sorry. Working it out over here, too. ^^;

23. anonymous

$2\sqrt{2}$...

24. anonymous

i keep getting wrong :(

25. anonymous

?

26. anonymous

$f(0) = \sqrt{8-6(0)^2} = \sqrt{8}= 2\sqrt{2}$

27. anonymous

I was actually getting kinda confused, and I ended up graphing it to check.