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Math Help - finding concavity

  1. #1
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    finding concavity

    Trying to find the points of inflection of f(x)= (1/4)x^4+(7/3)x^3-(5/2)x^2+2x.

    I got the intervals (1/3, 155/324) and (-5, -2495/12). This intervals seem a little extreme, did I compute this wrong.
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  2. #2
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    Quote Originally Posted by maddmaxx46 View Post
    Trying to find the points of inflection of f(x)= (1/4)x^4+(7/3)x^3-(5/2)x^2+2x.

    I got the intervals (1/3, 155/324) and (-5, -2495/12). This intervals seem a little extreme, did I compute this wrong.

    Perhaps: f'(x)=x^3+7x^2-5x+2\Longrightarrow\,f''(x)=3x^2+14x-5
    Thus the second derivative vanishes at x=-5\,,\,\,\frac{1}{3}

    As f''(x) is an upwards parabola, it is positive (and thus f(x) is concave upwards) at...

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Perhaps: f'(x)=x^3+7x^2-5x+2\Longrightarrow\,f''(x)=3x^2+14x-5
    Thus the second derivative vanishes at x=-5\,,\,\,\frac{1}{3}

    As f''(x) is an upwards parabola, it is positive (and thus f(x) is concave upwards) at...

    Tonio

    Ok but are my points of inflection correct? and there is a point where the graph concaves up and down. I plugged both answers i found for x into the orignal equation, that is how I found my points.
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