## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Section 14.7 - Extreme Values and saddle Points - Exercises 14.7 - Page 843: 26

#### Answer

Saddle point at $(0,0)$

#### Work Step by Step

Given: $f_x(x,y)=-y e^ x=0, f_y(x,y)=e^y-e^x=0$ Simplify the given two equations. This implies that, we have critical point: $(0, 0)$ In order to solve this problem we will have to apply Second derivative test that suggests the following conditions to calculate the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$. 1. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\gt 0$ , then $f(a,b)$ is a local minimum. 2. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\lt 0$ , then $f(a,b)$ is a local maximum. 3. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \lt 0$ , then $f(a,b)$ is a not a local minimum or local maximum but a saddle point. $D(0,0)=f_{xx}f_{yy}-f^2_{xy}=-1 \lt 0$ Thus, Saddle point at $(0,0)$

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