Inverse of a Function
Inverse function: g(x) = x − 3 — 2 x −11357 y −2 −1012 The graph of an inverse function is a refl ection of the graph of the original function The line of refl ection is y = x To fi nd the inverse of a function algebraically, switch the roles of x and y, and then solve for y Finding the Inverse of a Linear Function Find the inverse
A Guide to Functions and Inverses
7 Inverse of a Linear Function The method of getting the equation of an inverse of a linear function is discussed It is also given that the gradient would remain the same but the y-intercept would most
44 Solving Congruences using Inverses
inverse of a modulo m is congruent to a modulo m ) Proof By B ezout’s Theorem, since gcdpa;mq 1, there exist integers s and t such that 1 sa tm: Therefore sa tm 1 pmod mq: Because tm 0 pmod mq, it follows that sa 1 pmod mq: Therefore s is an inverse of a modulo m To show that the inverse of a is unique, suppose that there is another inverse
Math 133 Inverse Functions What is a function? Like other
Math 133 Inverse Functions Stewart x6 1 What is a function? Like other mathematical concepts, the idea of a function has four levels of meaning: physical (word problems), geometric (graphs), numerical (spread-sheets), and algebraic (formulas) We illustrate these with the following example Physical
Matrices, transposes, and inverses
Feb 01, 2012 · ⇒⇐ So it must be that case that the inverse of A is unique Take-home message: The inverse of a matrix A is unique, and we denote it A−1 Theorem (Properties of matrix inverse) (a) If A is invertible, then A −1is itself invertible and (A )−1 = A Lecture 7 Math 40, Spring ’12, Prof Kindred Page 2
Infinite Algebra 2 - Inverse Functions - Class Examples
Answers to Inverse Functions - Class Examples & Practice (ID: 1) 1) h-1 (x) = x + 2 3) g-1 (x) = 3-x - 2 2 5) h-1 (x) = 3 - 1 2 x 7) x y-6-4-2246-6-4-2 2 4 6 g-1 (x
MATH 1A - HOW TO SIMPLIFY INVERSE TRIG FORMULAS
MATH 1A - HOW TO SIMPLIFY INVERSE TRIG FORMULAS PEYAM RYAN TABRIZIAN Sample Problem (1 6 65) : Show cos(sin 1(x)) = p 1 x2 1 HOW TO WRITE OUT YOUR ANSWER Let = sin 1(x) (then sin( ) = x)
Inverse Matrices Date Period - Kuta Software LLC
no inverse Many answers Ex: 1 2 2 4 18) Give an example of a matrix which is its own inverse (that is, where A−1 = A) Many answers Ex: −10 9 −11 10-2-Create your own worksheets like this one with Infinite Algebra 2 Free trial available at KutaSoftware com
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Math 133Inverse FunctionsStewartx6.1
What is a function?Like other mathematical concepts, the idea of afunctionhas four levels of meaning: physical (word problems), geometric (graphs), numerical (spread- sheets), and algebraic (formulas). We illustrate these with the following example. Physical.We throw a stone upward o a 25 m building, with an initial vertical speed of 20 m/sec; and we assume gravitational acceleration of 10 m/sec2down-
ward. Letsbe the height of the stone in meters,tsec after the throw. Thus, the functions=f(t) is dened by looking to see the height at a given time. We limit our experiment tot2[0;5].Geometric.The stone's height is plotted approximately by:Numerical.We collect the following observations for height each second:t012345
s=f(t)25404540250 This is only a rough model: we could take a smaller increment for the time inputs, or indeed imagine an innite table to capture all inputs and outputs off(t). Algebraic:s=f(t) = 25+20t5t2. This agrees with the physical version because the initial height isf(0) = 25; the initial velocity isf0(0) = [2010t]t=0= 20; and the constant acceleration isf00(t) =10, negative meaning downward. This formula is only valid fort2[0;5], since the stone stops at the ground, and like all physical models, it is slightly thrown o by subtle factors such as air resistance. What is an inverse function?Continuing our example, we examine the inverse of the functions=f(t) above. This meansreversing the roles of input and output(the independent and dependent variables), so that timetbecomes a function of heights: in symbolst=f1(s). That is,f1is the rule that tells us at what timetthe stone reaches a given heights. Here we encounter a problem: since the stone rises and then falls, it reaches a given heightsattwodierent times: for example,f(1) =f(3) = 40. Thusf1(40) = 1 and/or3, with no single output, sof1is not a function. This happens because the original
functionfis notone-to-one: instead of taking dierent inputs to dierent outputs, ittakes two dierent inputst= 1;3 to the same outputs= 40. In graphical terms,Notes by Peter Magyarmagyar@math.msu.edu
s=f(t) fails thehorizontal line test, meaning a horizontal line likes= 40 intersects the graph more than once. To x this problem, we must restrict the domain of our function: we will look at the stone only fort2[2;5] (solid part of the graph). Technically, this denes a new function: f: [2;5]![0;45]; meaning the only allowed inputs aret2[2;5], which produce outputs covering the intervals2[0;45]. This restricted function is one-to-one, satisfying the horizontal line test, so we can get an inverse function: f1: [0;45]![2;5];
which again has four meanings: Physical.The inverse functiont=f1(s) gives the unique timet2[2;5] for which the stone is at heights. Geometric.The grapht=f1(s) is the original graphs=f(t) ipped diagonally,so as to switch the vertical and horizontal axes.Numerical.We restrict thes=f(t) table tot2[2;5], and switch the two rows so
thatsis the input row,tis the output row.s4540250 t=f1(s)2345 We can tidy this, rearranging and supplementing the data columns: s051015202530354045t=f1(s)5:04:84:64:44:24:03:73:43:02:0Algebraic:We must go froms= 25 + 20t5t2tot= formula ins. This just
means tosolvethe original equation fortin terms ofs. The Quadratic Formula solvesax2+bx+c= 0 asx=bpb24ac2a, so:
s= 25 + 20t5t2()(5)t2+ 20t+ (25s) = 0 ()t=20p2024(5)(25s)2(5)= 2q915
s: Thesign gives a choice between the two valuest1;t2in the original domain t2[0;5] which correspond tos. We want the larger choice, namely the + sign, to gett2[2;5] as required: t= 2 +q915 s: Note: In the relationshipss=f(t) andt=f1(s), the variablest;sare merely suggestive, recalling the physical meaning of these functions. On the algebraic level, we don't really care what the variables mean, and we sometimes change letters to makexthe input variable for every function: f(x) = 25 + 20x5x2; f1(x) = 2 +q915 x :Formal denitions.
Denition:Consider a functionf:A!Bwith inputs in the setAand outputs covering the setB. Supposefis one-to-one, meaning ifx16=x2, thenf(x1)6=f(x2). (The graphy=f(x) satises the horizontal line test.) Then we dene theinverse functionf1:B!Aasf1(b) =a, where a2Ais the unique value withf(a) =b. Inverse Theorem:The functionfand its inversef1satisfy: f1(f(a)) =aandf(f1(b)) =b
for alla2A,b2B. That is,fandf1undo each other under composition. The proof is just applying the denitions: iff(a) =b, thenf1(b) =a, which means f1(f(a)) =f1(b) =a; and similarly for the other equation.
In our previous example, this means:
f1(f(t)) = 2 +q915
(25+20t5t2) =t; f(f1(s)) = 25 + 202+q915
s 52+q915
s 2 =s: These equations are not obvious, but they must hold after simplication. Another Example.For the function:y=f(x) =x+1x+2, we can get its inverse by solving forxin terms ofy: y=x+1x+2()x+1 = (x+2)y ()x+1 =xy+ 2y ()xxy= 2y1 ()x(1y) = 2y1 ()x=f1(y) =2y11y: Changing the input variable tox, we get:f1(x) =2x11x. Note that the natural domain off(x), the set of inputs for which the formulax+1x+2 makes sense, is allx6=2. Also, we can writef(x) =x+21x+2= 11x+2, so the range of f(x), the set of all outputs, is ally6= 1. This is reversed for the inverse function:f1(y) has domainy6= 1 and rangex6= 2.Derivatives of inverse functions.
Inverse Derivative Theorem.Supposef:A!Bhas inversef1:B!A, thatf(a) =bfor some valuesa;b, and thatf(x) is dierentiable atx=a.Thenf1(y) is dierentiable aty=b:
(f1)0(b) =1f0(a)=1f
0(f1(b)):
In Leibnitz notation withy=f(x) andx=f1(y), we have: dxdy y=b=1dy dx x=a: Proof:This is most clear geometrically, considering that in the graphy=f(x), we havef0(a)yx, the rise-over-run for a small interval nearx=a. In the inverse graphx=f1(y), the vertical and horizontal increments (rise and run) are switched, so that (f1)0(b)xy1f0(a). Taking x;y!0 turns the approximations into exact
equalities in the limit.A dierent, algebraic proof comes from the equations in the Inverse Theorem. Taking
xas the input variable, we havef(f1(x)) =x. Applying the Chain Rule withf1(x) as the inside function gives: [f(f1(x))]0= (x)0=)f0(f1(x))(f1)0(x) = 1 =)(f1)0(x) =1f