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Math Help - Find the critical point(s) of f(x,y) = x + y^2 - e^x. Decide which are extreme points

  1. #1
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    Find the critical point(s) of f(x,y) = x + y^2 - e^x. Decide which are extreme points

    Find the critical point(s) of f(x,y) = x + y^2 - e^x. Decide which are extreme points and which are saddle points.

    I got to here:
    Px = 1 -e^x
    Py = 2y

    But now I'm stuck.

    I know I'm supposed to use this rule:

    d=fxxfyy-(fxy)^2

    If d>0:
    fxx(a,b)>0, then f(a, b) is a rel min
    fxx(a,b)<0, then f(a,b) is a rel. max

    If d<0, it's a saddle point.

    Please help! Thank you
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  2. #2
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    Quote Originally Posted by ewkimchi View Post
    Find the critical point(s) of f(x,y) = x + y^2 - e^x. Decide which are extreme points and which are saddle points.

    I got to here:
    Px = 1 -e^x
    Py = 2y

    But now I'm stuck.
    At least on this one you have made an effort! The whole point of taking partial derivatives is that the critical points occur where the derivatives are 0!

    Solve 1- e^x= 0 and 2y= 0.

    I know I'm supposed to use this rule:

    d=fxxfyy-(fxy)^2

    If d>0:
    fxx(a,b)>0, then f(a, b) is a rel min
    fxx(a,b)<0, then f(a,b) is a rel. max
    Yes, with a and b the values you got by solving those equations.

    If d<0, it's a saddle point.

    Please help! Thank you
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    At least on this one you have made an effort! The whole point of taking partial derivatives is that the critical points occur where the derivatives are 0!

    [snip]
    I hope that ! is an exclamation mark

    For the slow minded:

    Spoiler:
    0! = 1
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  4. #4
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    Thanks, but I what if fxx=0?

    Which in this case, it does
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    I hope that ! is an exclamation mark

    For the slow minded:

    Spoiler:
    0! = 1
    +1
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