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Math Help - extrema points

  1. #1
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    extrema points

    find absolute maximum of function f(x)= e^(-x)/(1+x^2)
    (a)1
    (b)2
    (3)e^-1
    (d)e^-1/2
    (e)none

    2.find minimum value of function f(x)=xlnx
    (a)-e
    (b)-1
    (c)-1/e
    (d)e^1/e
    (e)none
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bobby77 View Post
    find absolute maximum of function f(x)= e^(-x)/(1+x^2)
    (a)1
    (b)2
    (3)e^-1
    (d)e^-1/2
    (e)none
    Note: One of the nicest things to do is to graph the function, if you have that option available. A simple look at the graph will tell you that the answer is e). Failing that possibility:

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

    f'(x) = \frac{-e^{-x}(1 + x^2) - e^{-x}(2x)}{(1+x^2)^2} by the product rule.

    We need to find where this is zero.
    0 = \frac{-e^{-x}(1 + x^2) - e^{-x}(2x)}{(1+x^2)^2}

    We only need to numerator to be 0, so:
    0 = -e^{-x}(1 + x^2) - e^{-x}(2x) Divide both sides by e^{-x} (which is never 0):

    0 = -(1 + x^2) - 2x

    Now solve for x:
    -x^2 - 2x - 1 = 0

    Thus x = -1. (Note: this value of x does NOT make the denominator 0.)

    Now, the point (-1, f(-1)) is merely a critical point. It may be a max or a min and it may only be local. It may also be an inflection point. We need to check all these possibilities.

    First comment: The function f(x) has no vertical asymptotes.
    A reasonable check on whether the critical point is a local or absolute maximum (in the absence of vertical asymptotes) is to take the limit of the function as x goes to positive and negative infinity to see if it "blows up" or "blows down."

    \lim_{x \to \infty}\frac{e^{-x}}{1+x^2} = 0
    \lim_{x \to -\infty}\frac{e^{-x}}{1+x^2} \to \infty

    Since your function goes to infinity as x approaches to negative infinity, your critical point could only possibly be a local max at best. (As it happens, x = -1 is actually an inflection point, not a local max or min.)

    Thus the function has no absolute maximum.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    I suppose I should finish the problem, even though we have enough information to answer the question.

    How do we determine that x = -1 provides an inflection point?

    The next step to go to is the second derivative test. The first derivative of f(x) is:

    f'(x) = \frac{-e^{-x}(1+x^2) - e^{-x}(2x)}{(1+x^2)^2} = - \frac{e^{-x}(x+1)^2}{(1+x^2)^2}

    So
    f''(x) = - \frac{ \left ( -e^{-x}(x+1)^2 + e^{-x}2(x+1) \right )(1+x^2)^2 - e^{-x}(x+1)^2 \cdot 2(1+x^2)2x}{(1+x^2)^4}

    If f''(x) > 0 at x = -1 then x = -1 represents a local minimum for the function.
    If f''(x) < 0 at x = -1 then x = -1 represents a local maximum for the function.
    If f''(x) = 0 at x = -1 then x = -1 represents an inflection point of the function.

    Since f''(-1) = 0, x = -1 is an inflection point.

    -Dan
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  4. #4
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    Quote Originally Posted by topsquark View Post
    Note: One of the nicest things to do is to graph the function, if you have that option available. A simple look at the graph will tell you that the answer is e). Failing that possibility:

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

    f'(x) = \frac{-e^{-x}(1 + x^2) - e^{-x}(2x)}{(1+x^2)^2} by the product rule.

    We need to find where this is zero.
    0 = \frac{-e^{-x}(1 + x^2) - e^{-x}(2x)}{(1+x^2)^2}

    We only need to numerator to be 0, so:
    0 = -e^{-x}(1 + x^2) - e^{-x}(2x) Divide both sides by e^{-x} (which is never 0):

    0 = -(1 + x^2) - 2x

    Now solve for x:
    -x^2 - 2x - 1 = 0

    Thus x = -1. (Note: this value of x does NOT make the denominator 0.)

    Now, the point (-1, f(-1)) is merely a critical point. It may be a max or a min and it may only be local. It may also be an inflection point. We need to check all these possibilities.

    First comment: The function f(x) has no vertical asymptotes.
    A reasonable check on whether the critical point is a local or absolute maximum (in the absence of vertical asymptotes) is to take the limit of the function as x goes to positive and negative infinity to see if it "blows up" or "blows down."

    \lim_{x \to \infty}\frac{e^{-x}}{1+x^2} = 0
    \lim_{x \to -\infty}\frac{e^{-x}}{1+x^2} \to \infty

    Since your function goes to infinity as x approaches to negative infinity, your critical point could only possibly be a local max at best. (As it happens, x = -1 is actually an inflection point, not a local max or min.)

    Thus the function has no absolute maximum.

    -Dan
    There is no need to look for stationary points for this function, as x-> -infty
    the e^(-x) term goes to infinity, and faster than any polynomial in x, so the
    whole thing goes to infinity as x->-infty.

    RonL
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  5. #5
    Forum Admin topsquark's Avatar
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    I suppose I should finish the problem, even though we have enough information to answer the question.

    How do we determine that x = -1 provides an inflection point?

    The next step to go to is the second derivative test. The first derivative of f(x) is:

    f'(x) = \frac{-e^{-x}(1+x^2) - e^{-x}(2x)}{(1+x^2)^2} = - \frac{e^{-x}(x+1)^2}{(1+x^2)^2}

    So
    f''(x) = - \frac{(-e^{-x}(x+1)^2 + e^{-x}2(x+1))(1+x^2)^2 - e^{-x}(x+1)^2 \cdot 2(1+x^2)2x}{(1+x^2)^4}

    If f''(x) > 0 at x = -1 then x = -1 represents a local minimum for the function.
    If f''(x) < 0 at x = -1 then x = -1 represents a local maximum for the function.
    If f''(x) = 0 at x = -1 then x = -1 represents an inflection point of the function.

    Since f''(-1) = 0, x = -1 is an inflection point.

    -Dan
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by bobby77 View Post
    2.find minimum value of function f(x)=xlnx
    (a)-e
    (b)-1
    (c)-1/e
    (d)e^1/e
    (e)none
    Now this one has its global minimum at a stationary point.

    <br />
\frac{df}{dx}=\ln(x)+1<br />

    so the stationary points are the solutions of:

    <br />
\frac{df}{dx}=\ln(x)+1=0<br />

    which has a single solution at \ln(x)=-1, or x=e^{-1}, where f(x)=-1/e

    RonL
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