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

1. MoonlitFate

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

2. mathcalculus

yes

3. mathcalculus

i mean x and y

4. mathcalculus

critical points..

5. MoonlitFate

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

6. mathcalculus

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

7. MoonlitFate

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

8. MoonlitFate

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. MoonlitFate

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

10. mathcalculus

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

11. mathcalculus

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. MoonlitFate

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

13. MoonlitFate

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. mathcalculus

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

15. mathcalculus

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

16. mathcalculus

17. MoonlitFate

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

18. mathcalculus

its okay can we do after we find the x

19. mathcalculus

the y..

20. MoonlitFate

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

21. mathcalculus

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

22. MoonlitFate

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

23. mathcalculus

$2\sqrt{2}$...

24. mathcalculus

i keep getting wrong :(

25. mathcalculus

?

26. MoonlitFate

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

27. MoonlitFate

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