Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Practice Exercises - Page 866: 70


Saddle Point of $f(0,1)=-3$ Local Minimum $f(2, 1)=-19$ Local Minimum $f(-2, 1)=-19$

Work Step by Step

$$f_x(x,y)=4x^3-16x=0$$ and $$f_y(x,y)=6y-6=0$$ Critical points are: $(-2,1 )$ and $(0,1)$ and $(2,1)$ Apply the second derivative test for $(-2,1 )$ and $(0,1)$ and $(2,1)$ $D=f_{xx}f_{yy}-f^2_{xy}=(-16)(6)-0=-96-0=-96 \lt 0$ $D(2,1)=f_{xx}f_{yy}-f^2_{xy}=(-16)(6)-0^2=192-0=192 \gt 0$ $D(-2,1) =f_{xx}f_{yy}-f^2_{xy}=(32)(6)-0^2=192-0=192 \gt 0$ So, Local Minimum $f(-2, 1)=-19$ Local Minimum $f(2, 1)=-19$ Saddle Point of $f(0,1)=-3$
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