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Thread: Range of Functions

  1. #1
    Senior Member harpazo's Avatar
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    Range of Functions

    Find the range of f(x) = 2x^2 - x - 1.

    How is this done algebraically?
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    MHF Contributor MarkFL's Avatar
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    Re: Range of Functions

    We observe that the given function is a quadratic, whose graph opens upwards since the coefficient on the squared term is positive. So, we know there is no upper bound on the range. Thus the lower bound will occur on the axis of symmetry:

    $\displaystyle x=-\frac{-1}{2(2)}=\frac{1}{4}$

    $\displaystyle f_{\min}=f\left(\frac{1}{4}\right)=?$
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  3. #3
    Senior Member harpazo's Avatar
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    Re: Range of Functions

    Quote Originally Posted by MarkFL View Post
    We observe that the given function is a quadratic, whose graph opens upwards since the coefficient on the squared term is positive. So, we know there is no upper bound on the range. Thus the lower bound will occur on the axis of symmetry:

    $\displaystyle x=-\frac{-1}{2(2)}=\frac{1}{4}$

    $\displaystyle f_{\min}=f\left(\frac{1}{4}\right)=?$
    What do you mean by upper and lower bound?
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    MHF Contributor MarkFL's Avatar
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    Re: Range of Functions

    Quote Originally Posted by harpazo View Post
    What do you mean by upper and lower bound?
    The global maximum/minimum.
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  5. #5
    Senior Member harpazo's Avatar
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    Re: Range of Functions

    Quote Originally Posted by MarkFL View Post
    The global maximum/minimum.
    Can you show me the meaning of upper and lower bound using a graph?
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    MHF Contributor MarkFL's Avatar
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    Re: Range of Functions

    Quote Originally Posted by harpazo View Post
    Can you show me the meaning of upper and lower bound using a graph?
    For a parabola opening upwards, the minimum is at the vertex, and there is no maximum as the function has no upper bound, that is, it can be made as large as we want the farther we take an x value from the axis of symmetry.
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  7. #7
    Senior Member harpazo's Avatar
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    Re: Range of Functions

    Quote Originally Posted by MarkFL View Post
    For a parabola opening upwards, the minimum is at the vertex, and there is no maximum as the function has no upper bound, that is, it can be made as large as we want the farther we take an x value from the axis of symmetry.
    Ok. If the coefficient of the x^2 term is positive, the parabola opens upward. Thus, we have a minimum at the vertex.

    If the coefficient of the x^2 term is negative, the parabola opens downward. Thus, we have a maximum at the vertex.

    Is this right?
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: Range of Functions

    Quote Originally Posted by harpazo View Post
    Ok. If the coefficient of the x^2 term is positive, the parabola opens upward. Thus, we have a minimum at the vertex.

    If the coefficient of the x^2 term is negative, the parabola opens downward. Thus, we have a maximum at the vertex.

    Is this right?
    Yes, that's correct.
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  9. #9
    Senior Member harpazo's Avatar
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    Re: Range of Functions

    Quote Originally Posted by MarkFL View Post
    Yes, that's correct.
    Ok. I have a somewhat understanding of max/min.
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