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Find all the relative extrema (local extrema) and the saddle points of the function f(x, y) = (y² +1)x³ — (x − 1)³. Solution Step 1. Find all the critical points of the function f(x, y). We have f₂ = fy fz = 0 Solve fy = 0' The partial derivatives fr, fy exist for all (x, y) ≤ R². we obtain that M At M, D= Step 2. Apply the second partial derivatives test to verify if f has a relative extremum at M or not. We have A = frz = D = fzzfyy - fy is a critical point of the function f(x,y). 0, A 0, that implies f(x,y) + at the point M.

Find all the relative extrema (local extrema) and the saddle points of the function f(x, y) = (y² +1)x³ — (x − 1)³. Solution Step 1. Find all the critical points of the function f(x, y). We have f₂ = fy fz = 0 Solve fy = 0' The partial derivatives fr, fy exist for all (x, y) ≤ R². we obtain that M At M, D= Step 2. Apply the second partial derivatives test to verify if f has a relative extremum at M or not. We have A = frz = D = fzzfyy - fy is a critical point of the function f(x,y). 0, A 0, that implies f(x,y) + at the point M.

Calculus: Early Transcendentals
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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Chapter 11. Partial derivatives

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Find all the relative extrema (local extrema) and the saddle points of the function f(x, y) = (y² + 1)x³ − (x − 1)³. Solution Step 1. Find all the critical points of the function f(x, y). We have fx fy S fa | fy The partial derivatives fa, fy exist for all (x, y) = R². Solve fx = 0 At M, D= = 0 we obtain that M Step 2. Apply the second partial derivatives test to verify if f has a relative extremum at M or not. We have A = fxx D = fax fyy - fay is a critical point of the function f(x,y). 0, A= 0, that implies f(x,y) at the point M.
Transcribed Image Text:Find all the relative extrema (local extrema) and the saddle points of the function f(x, y) = (y² + 1)x³ − (x − 1)³. Solution Step 1. Find all the critical points of the function f(x, y). We have fx fy S fa | fy The partial derivatives fa, fy exist for all (x, y) = R². Solve fx = 0 At M, D= = 0 we obtain that M Step 2. Apply the second partial derivatives test to verify if f has a relative extremum at M or not. We have A = fxx D = fax fyy - fay is a critical point of the function f(x,y). 0, A= 0, that implies f(x,y) at the point M.
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