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Homework on 30 October 2017:

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  1. In single-variable calculus, you used the second derivative to determine whether a critical point of a function was a maximum or a minimum. For multivariable calculus, we also have a second-derivative test, though it is more complicated.

    Second-Derivative Test: Suppose $f\colon\R^2\to\R$ has continuous second partial derivatives at a critical point $(x_0,y_0)$ of $f$. Let $$ D = f_{xx}(x_0,y_0)f_{yy}(x_0,y_0) - \bigl(f_{xy}(x_0,y_0)\bigr)^2 $$ Then:

    1. If $D > 0$ and $f_{xx}(x_0,y_0) > 0$ then $f$ has a relative minimum at $(x_0,y_0)$.
    2. If $D > 0$ and $f_{xx}(x_0,y_0) < 0$ then $f$ has a relative maximum at $(x_0,y_0)$.
    3. If $D < 0$ then $f$ has a saddle point at $(x_0,y_0)$.
    4. If $D = 0$ then the test tells us nothing (no conclusion can be drawn).

    Consider the function $f(x,y) = xy - x^3 - y^2$.

    1. Locate all the critical points of $f$.

    2. Use the second derivative test to determine the type of critical point in each case.

  2. Show that the fourth condition above is necessary by producing examples of functions where $(0,0)$ is a critical point and $D=0$ at $(0,0)$ but where:

    1. The function has a minimum at $(0,0)$.
    2. The function has a maximum $(0,0)$.
    3. The function has a saddle $(0,0)$.
    4. The function has none of the above $(0,0)$.
    (Hint: Try to do something like our single-variable examples from class.)

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Created: 30 Oct 2017
Last modified: Oct 30, 2017 8:01:53 PM
Comments to: dpvc@union.edu
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