Let f:Rx R→ Rx R be given by f(a,b) = (3a + 2b,2a –b) a, b) ERx R. 1) Find f(2,3) and f(– 1,4). 2) Construct an argument to determine if f is an injection using the definition of injection. That is, adapt the definition of injective for a function of a single variable: f is injective if f(a)=f(b) implies a = b. Hint: Check the textbook. 3Let (r s) be an arhitrany element of the codomain of f Work bac dh Verify that for that for these values of a and b f(a b)=(r s) Can we conclude that f je a hijection?

Let f:Rx R→ Rx R be given by f(a,b) = (3a + 2b,2a –b) a, b) ERx R. 1) Find f(2,3) and f(– 1,4). 2) Construct an argument to determine if f is an injection using the definition of injection. That is, adapt the definition of injective for a function of a single variable: f is injective if f(a)=f(b) implies a = b. Hint: Check the textbook. 3Let (r s) be an arhitrany element of the codomain of f Work bac dh Verify that for that for these values of a and b f(a b)=(r s) Can we conclude that f je a hijection?

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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Question

Let f:Rx R→ Rx R be given by f(a,b) = (3a + 2b,2a – b) V(a, b) ERx R.
1) Find f(2,3) and f(- 1,4).
2) Construct an argument to determine if f is an injection using the definition of injection. That is, adapt the definition of injective for a function of a single variable: f is injective if f(a) = f(b) implies a = b. Hint: Check the
textbook.
3) Let (r,s) be an arbitrary element of the codomain of f. Work backwards to find a and b. Verify that for that for these values of a and b, f(a,b) = (r, s). Can we conclude that f is a bijection?

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