Answered: Find the critical points and determine…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

Find the critical points and determine if it is a maximum, minimum or saddle point. value, minimum or saddle point.

please, just do this exercise, I have this question, I am a high school student

I. Encuentre los puntos críticos y determine si es valor máximo, mínimo o
punto silla.
f(x,y) = (x² + y²)evi-x²

Expert Solution

Step 1: Given and basic information

$G i v e n : f (x, y) = (x^{2} + y^{2}) e^{(y^{2} - x^{2})}$

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$F o r m u l a e u s e d : \frac{d}{d x} f (x) g (x) = f' (x) g (x) + f (x) g' (x) \frac{d}{d x} e x p (g (x)) = e x p (g (x)) \cdot g' (x) \frac{d}{d x} (a x^{n}) = a n x^{(n - 1)}$

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$T e s t t o d e t e r m i n e l o c a l e x t r e m a a n d s a d d l e p o i n t : D (x, y) = \frac{\partial^{2}}{\partial x^{2}} f (x, y) \frac{\partial^{2}}{\partial y^{2}} f (x, y) - {(\frac{\partial^{2}}{\partial x \partial y} f (x, y))}^{2} I f T e s t R e m a r k s D (x, y) < 0 (S a d d l e p o i n t a t (x, y)) D (x, y) = 0 (I n c o n c l u s i v e) D (x, y) > 0 & \frac{\partial^{2}}{\partial x^{2}} f (x, y) < 0 (l o c a l m a x i m a a t (x, y)) D (x, y) > 0 & \frac{\partial^{2}}{\partial x^{2}} f (x, y) > 0 (l o c a l m i n i m a a t (x, y))$

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$f (x, y) = (x^{2} + y^{2}) e^{(y^{2} - x^{2})} (1) \frac{\partial}{\partial x} f (x, y) = e^{(y^{2} - x^{2})} (2 x) + (x^{2} + y^{2}) e^{(y^{2} - x^{2})} (- 2 x) = e^{(y^{2} - x^{2})} (2 x - 2 x^{3} - 2 x y^{2}) (2) \frac{\partial}{\partial y} f (x, y) = e^{(y^{2} - x^{2})} (2 y) + (x^{2} + y^{2}) e^{(y^{2} - x^{2})} (2 y) = e^{(y^{2} - x^{2})} (2 y + 2 y x^{2} + 2 y^{3}) (3)$

$\frac{\partial^{2}}{\partial x^{2}} f (x, y) = - 2 x e^{(y^{2} - x^{2})} (2 x - 2 x^{3} - 2 x y^{2}) + e^{(y^{2} - x^{2})} (2 - 6 x^{2} - 2 y^{2}) = e^{(y^{2} - x^{2})} (4 x^{4} + 4 x^{2} y^{2} + 2 - 10 x^{2} - 2 y^{2}) (4) \frac{\partial^{2}}{\partial x \partial y} f (x, y) = e^{(y^{2} - x^{2})} (2 y (2 x - 2 x^{3} - 2 x y^{2}) + (- 4 x y)) = e^{(y^{2} - x^{2})} (4 y x^{3} - 4 x y^{3}) (5) \frac{\partial^{2}}{\partial y^{2}} f (x, y) = e^{(y^{2} - x^{2})} (2 y (2 y + 2 y x^{2} + 2 y^{3}) + (2 + 2 x^{2} + 6 y^{2})) = e^{(y^{2} - x^{2})} (10 y^{2} + 4 y^{2} x^{2} + 4 y^{4} + 2 + 2 x^{2}) (6)$