Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - 14.7 Exercises - Page 978: 20

Answer

See explanation below for the proof.

Work Step by Step

Second derivative test: Some noteworthy points to calculate the local minimum, local maximum and saddle point of $f$. 1. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum. 2.If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum. 3. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is not a local minimum and local maximum or, a saddle point. As we are given that $f(x,y)=x^2ye^{-x^2-y^2}$ For $(x,y)=( 1,\dfrac{1}{\sqrt 2})$ $D=8e^{-3} \gt 0$ ; and $f_{xx} \lt 0$ For $(x,y)=( -1,\dfrac{1}{\sqrt 2})$ $D=8e^{-3} \gt 0$ ; and $f_{xx} \lt 0$ For $(x,y)=( 1,-\dfrac{1}{\sqrt 2})$ $D=8e^{-3} \gt 0$ ; and $f_{xx} \gt 0$ For $(x,y)=( -1,-\dfrac{1}{\sqrt 2})$ $D=8e^{-3} \gt 0$ ; and $f_{xx} \gt 0$ $D(0,0) \lt 0$; saddle points; because when $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is not a local minimum and local maximum or, a saddle point. Also, $f_{xx}=2y,f_{yy}=0, f_{xy}=0$ and $f_{xx}(x,y)f_{yy}(x,y)-[f_{xy}(x,y)]^2 = 0$ Therefore, we have Maximum value: $f(\pm 1, \dfrac{1}{\sqrt 2})$, Minimum value: $f(\pm 1, -\dfrac{1}{\sqrt 2})$ and saddle point:$(0, 0)$. This implies that $D=0$ that is, the function $f$ has infinite critical points along the line $x=0$ on the line $x=0$ at the points $f(\pm 1, \pm \dfrac{1}{\sqrt 2})$ Hence, the result has been proved.
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