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Math Help - Help for a function please

  1. #1
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    Help for a function please

    I don't have the slightest idea how to find the turning points of this function; f(x) = 2/3X^3 - 1/2X^2 + 4
    Could you please help me?
    I'd appreciate it a lot
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  2. #2
    MHF Contributor
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    Quote Originally Posted by FionaG View Post
    I don't have the slightest idea how to find the turning points of this function; f(x) = 2/3X^3 - 1/2X^2 + 4
    Could you please help me?
    I'd appreciate it a lot
    You have to differentiate that function and have it equal 0 .

     f(x)=\frac{2}{3}x^3-\frac{1}{2}x^2+4

     \frac{dy}{dx}=2x^2-x

     2x^2-x=0 , find the values of x and sub them into the original equation to get the y's .
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  3. #3
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    Thanks, but could anyone tell me the result? I just don't understand this one
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  4. #4
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    x(2x-1)=0\rightarrow x=0\text{ or }\frac 12\Rightarrow y=4\text{ or }\frac 23\cdot\left(\frac 12\right)^3-\frac 12\cdot \left(\frac 12\right)^3+4\approx 3.958 . So the points are: (0|4), \left(\frac 14|\approx 3.958\right).
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