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

1. MoonlitFate Group Title

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

2. mathcalculus Group Title

yes

3. mathcalculus Group Title

i mean x and y

4. mathcalculus Group Title

critical points..

5. MoonlitFate Group Title

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

6. mathcalculus Group Title

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

7. MoonlitFate Group Title

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

8. MoonlitFate Group Title

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 Group Title

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

10. mathcalculus Group Title

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

11. mathcalculus Group Title

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 Group Title

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

13. MoonlitFate Group Title

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 Group Title

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

15. mathcalculus Group Title

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

16. mathcalculus Group Title

17. MoonlitFate Group Title

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

18. mathcalculus Group Title

its okay can we do after we find the x

19. mathcalculus Group Title

the y..

20. MoonlitFate Group Title

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

21. mathcalculus Group Title

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

22. MoonlitFate Group Title

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

23. mathcalculus Group Title

$2\sqrt{2}$...

24. mathcalculus Group Title

i keep getting wrong :(

25. mathcalculus Group Title

?

26. MoonlitFate Group Title

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

27. MoonlitFate Group Title

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