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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) .

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  1. MoonlitFate
    • 2 years ago
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    So the function h(x) = \[\sqrt{8-6x^2}\] , is that right? By critical points, do you mean relative maxima / minima?

  2. mathcalculus
    • 2 years ago
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    yes

  3. mathcalculus
    • 2 years ago
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    i mean x and y

  4. mathcalculus
    • 2 years ago
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    critical points..

  5. MoonlitFate
    • 2 years ago
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    Oh, okay. First of all, do you know how to take the derivative of h(x)?

  6. mathcalculus
    • 2 years ago
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    so far im here: -12x/ 2 radical 8-6x^2

  7. MoonlitFate
    • 2 years ago
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    The derivative of h(x) is: \[\frac{ -6x }{ \sqrt{8-6x ^{2}} }\]

  8. MoonlitFate
    • 2 years ago
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    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
    • 2 years ago
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    So, you're going to set f'(x) to 0 and solve for x.

  10. mathcalculus
    • 2 years ago
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    okay makes sense... then to find the critical points i equal that to zero... and just solve for x?

  11. mathcalculus
    • 2 years ago
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    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
    • 2 years ago
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    \[\frac{ -6x }{ \sqrt{8-6x ^{2}} } = 0 \] \[(\sqrt{8-6x ^{2}}) *\frac{ -6x }{ \sqrt{8-6x ^{2}} } = 0 \sqrt{8-6x ^{2}}\] \[-6x = 0\] \[x=0\]

  13. MoonlitFate
    • 2 years ago
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    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
    • 2 years ago
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    I got zero as well... not how do i find y

  15. mathcalculus
    • 2 years ago
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    i plugged in zero into the normal question... and got 2 radical 2

  16. mathcalculus
    • 2 years ago
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    answer: (0,2 radical 2) is wrong://

  17. MoonlitFate
    • 2 years ago
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    Gah, I was doing something more complicated. Ignore the part about test intervals, sorry. :(

  18. mathcalculus
    • 2 years ago
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    its okay can we do after we find the x

  19. mathcalculus
    • 2 years ago
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    the y..

  20. MoonlitFate
    • 2 years ago
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    Plug in the critical numbers into the f(x).

  21. mathcalculus
    • 2 years ago
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    i plug in 0 in the original function? to find y right

  22. MoonlitFate
    • 2 years ago
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    Yes, sorry. Working it out over here, too. ^^;

  23. mathcalculus
    • 2 years ago
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    \[2\sqrt{2}\]...

  24. mathcalculus
    • 2 years ago
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    i keep getting wrong :(

  25. mathcalculus
    • 2 years ago
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    ?

  26. MoonlitFate
    • 2 years ago
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    \[f(0) = \sqrt{8-6(0)^2} = \sqrt{8}= 2\sqrt{2}\]

  27. MoonlitFate
    • 2 years ago
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    I was actually getting kinda confused, and I ended up graphing it to check.

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