If f(x, y) = x' – 2xy + 3y², find: (a) 2 f(x, y + k) – f(x, y) k + 0. 6.1. (c) х у k (a) ƒ(-2,3) = (-2)³ – 2(-2)(3) + 3(3)² = –8 + 12 + 27 = 31 1 2 (b) f 12 ху у f(x, y+ k) – f(x, y) _ 1 {[x³ – 2x(y+ k)+3(y+ k)²]–[x³ – 2xy+ 3y²]} k - 2.xy – 2kx + 3y + 6ky + 3k² – x² + 2.xy – 3y²) k 1 (-2kr = 6ky + 3k?) = -2x+6v+3k. 4,
If f(x, y) = x' – 2xy + 3y², find: (a) 2 f(x, y + k) – f(x, y) k + 0. 6.1. (c) х у k (a) ƒ(-2,3) = (-2)³ – 2(-2)(3) + 3(3)² = –8 + 12 + 27 = 31 1 2 (b) f 12 ху у f(x, y+ k) – f(x, y) _ 1 {[x³ – 2x(y+ k)+3(y+ k)²]–[x³ – 2xy+ 3y²]} k - 2.xy – 2kx + 3y + 6ky + 3k² – x² + 2.xy – 3y²) k 1 (-2kr = 6ky + 3k?) = -2x+6v+3k. 4,
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
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A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
Question
6.1) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text.
![Functions and graphs
Iff(x, y) - x* - 2ху + Зу?, find: (a) f
f(x, у + k) — F(x, у) k+0.
(с)
6.1.
k
(а) f(-2,3) %3 (-2)* — 2(-2)(3) + 3(3)? %—D —8 + 12 + 27%3 31
3
12
+
y?
1
4
(b) f
%3D
x²
ху
f(х, у + k) - f(x, у) _1
- (x - 2x(у + k) + 3(у+ k)*]-[x* - 2ху + Зу?]}
k
(с)
k
1
(х3 — 2ху - 2kx + 3Зу? + 6ky + 3k? - х? + 2ху - Зу?)
(-2kx —D 6ky + 3k') — —2х + бу + 3к.
k
=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad530511-9758-42d9-9994-894bc4eea3f3%2F804cba28-a552-4e4b-ad05-56ae0f9ca289%2Fnvosv6p_processed.png&w=3840&q=75)
Transcribed Image Text:Functions and graphs
Iff(x, y) - x* - 2ху + Зу?, find: (a) f
f(x, у + k) — F(x, у) k+0.
(с)
6.1.
k
(а) f(-2,3) %3 (-2)* — 2(-2)(3) + 3(3)? %—D —8 + 12 + 27%3 31
3
12
+
y?
1
4
(b) f
%3D
x²
ху
f(х, у + k) - f(x, у) _1
- (x - 2x(у + k) + 3(у+ k)*]-[x* - 2ху + Зу?]}
k
(с)
k
1
(х3 — 2ху - 2kx + 3Зу? + 6ky + 3k? - х? + 2ху - Зу?)
(-2kx —D 6ky + 3k') — —2х + бу + 3к.
k
=
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