MAT135H1 Study Guide - Quiz Guide: Quadratic Function, Product Rule, Graph Of A Function

Consider the set X, and let F = {f|j: X rightarrow X), the set of functions from X to itself. A notable member of F is, of course, the identity function, id_x: X rightarrow X, defined by the formula: Forall x Element X. id_x(x) = x. In this exercise, we will study the concept of the inverse function. Recall that given two functions f, g Element F we have defined the operation of composition, which is denoted o and takes two functions f, g Element F and produces a new function as follows: f og(x) = f(g(x) for all x Element X. We will give the following definitions: (i) Def 1. A function g Element F if called an left inverse of a function f Element F go f = id_x. (ii) Def 2. A function h Element F if called an right inverse of a function f Element F if f o h = id_x. (iii) Def 3. A function l Element F if called an inverse of a function f Element F if l is simultaneously a left and a right inverse of f. Let LI(f) the predicate "f has a left inverse", RI(f) the predicate "f has a right inverse", and INV(f) the predicate "f has an inverse". Also let INJ(f) be the predicate "f is injective (one-to-one)", SUR(f) "f is surjective (onto)", and "BIJ(f) "f is bijective (one-to-one and onto). Express each of the following statements in predicate logic and write a proof. (i) A function f: X rightarrow X is injective if and only if it has a left inverse. (ii) A function f: X rightarrow X is surjective if and only if it has a right inverse. (iii) A function f: X rightarrow X is bijective if and only if it has an inverse. (iv) If the inverse g of a function f: X rightarrow X exists, it is unique. ("Unique" means the following: if f has another inverse g', then g = g'. Since the inverse of a function is unique, there is a special notation used for it: f ^-1).

QUESTION 3 Question When using the method of Lagrange multipliers ▽f = λ▽g + μ▽h to find the extrema of the function f = (4)x2 + (2), (I)22 subject to the constraints 0-g = r +y2-r-1 and 0 = h = (x-2)2 + (v-2)2-1, one has to solve a system of three equations for variables r.) There is exactly one such solution that satisfies z > 0 and has 16.22108. The value ofr +y + z + ג for that solution is (A) 9.90535 (B) 2.24411 (C) 1.32064 (D) 3.685 (E) 7.0194 (F) 4.71473 (G) 8.00271 (H) 2.77668 B 0C. @D.@E.

Let r in and u =(1 1 0), v = (1 4 1), w = (r2 1 r2), b = (3+2r 5+12r 2r). For which values r in is the set {u. v, w} linearly independent? For which values r in is the vector b a linear combination of u. v and w? For which of these values of r can b be written as a linear combination of u. v and w in more than one way? The set 3 of all column vectors of length three, with real entries, is a vector space. Is the subset B = {(x y z) in 3 | xy + yz = 0} a subspace of 3? Justify your answer. Show that the set of all twice differentiable functions f : satisfying the differential equation sin(x) f"(x) + x2f(x) = 0 is a vector space with respect to the usual operations of addition of functions and multiplication by scalars. Here, f" denotes the second derivative of f. Let S be the following subset of the vector space 3 of all real polynomials p of degree at most 3: S = {p in 3 | p(l) = 0, p'(l) = 0}, where p' is the derivative of p. Determine whether S is a subspace of 3. Determine whether the polynomial q(x) = x - 2x2 + x3 is an element of S. u = (1 1 0), v = (1 4 1), w = (r2 1 r2) if {u, v, w} is linearly independent set 1(4r2 - 1)-1(r2 - r2) ne 0 4r2 ne 1 r ne plusmn1/2 this set is linearly independent set for values of e ne plusmn1/2 - - - - - - - (3+2r 5+12r 2r) = a1 (1 1 0) + a2 (1 4 1) + a3 (r2 1 r2) this forms system of equations in a1, a2, a3 if 4r2 - 1 ne 0 then system has unique solution if 4r2 - 1 = 0 and -2 - 4r = 0 rArr r = -1/2, the system has more than one solution if 4r2 - 1 ne 0 b can be written as linear combination of u,v,w as uniquely if r = -1/2 then b can be written as linear combination of u,v,w more than one way - - - - - - - - - (1 1 -1) in B and (1 0 1) in B (1 1 -1) + (1 0 1) = (2 1 0) not in B since 2times1+1times0 ne 0 so addition of vector is failed, this set is not forms subspace consider the following differential equation (sinx)f"(x) + x2f (x) = 0 each solution is continuous function the set of all solutions is subset of space of set of all continuous functions n from R to R let f,g are two solutions (sinx)f"(x) + x2f (x) = 0 (sinx)g"(x)+x2g(x) = 0 ccheck whether af+bg is solution or not (sinx)f"(x) + x2f(x) = 0 rArr (sinx) af"(x) + x2?f (x) = 0 (sinx)g"(x) + x2g (x) = 0 rArr (sinx)bg" (x) + x2bg (x) = 0 (sinx)(af"(x)+bg"(x)) +x2 (af (x) + bg(x)) = 0 af+bg is also solution to differential equation the set of all solutions to differential equation forms subspace - - - - - - - - - - consider the following subset of P3 {p in P3/.p(l)=0, p'(l) = 0} let p, q in S rArr p(l)= 0, p'(l) = 0 and q(l) = 0, q'(l) = 0 (ap + bq) (l) = ap (l) +b q(l) = a (0) + b (0) = 0 (ap + bq)'(1) = ap'(l) +bq'(1) = a (0) + b(0) = 0 p, q in S rArr ap +bq in S then S is subspace q(x) = x - 2x2 + x3 q(1) = 1 - 2 + 1 = 0 q'(x)=1 - 4x + 3x2 q'(l) = 1 - 4 + 3 = 0 q is element of S