1 Solutions to the in class part

Transcription

1 NAME: Solutions to te in class part. Te grap of a function f is given. Calculus wit Analytic Geometry I Exam, Friday, August 30, 0 SOLUTIONS (a) State te value of f(). (b) Estimate te value of f( ). (c) For wat values of x is f(x) =? (d) Estimate te value of x suc tat f(x) = 0. (e) State te domain and range of f. (f) On wat interval is f increasing. (a) f() = 3. (b) f( ) /3 (any value in te range 0.4 to 0. is acceptable.) (c) For x = and x = 3. (d) f( ) /3 (any value in te range 0.75 to 0.6 is acceptable.) (e) Te domain is (, 4). Te range is (, ]. (f) On te interval (, ].. If f(x) = x 3 f(a + ) f(a), evaluate. Your final expression sould be as simple as possible. Tat is, anyting tat can be canceled or simplified in any way, sould be canceled and simplified.

2 SOLUTIONS TO THE TAKE HOME PART f(a + ) f(a) = (a + )3 a 3 = 3a + 3a + 3 = 3a + 3a +. = a3 + 3a + 3a + 3 a 3 = (3a + 3a + ) = a3 + 3a + 3a + 3 a 3 = (3a + 3a + ) 3. Find te domain of te function f(x) = x3 5 x + x 6. Te only problem tis function as is tat te denominator can be 0. Te quadratic equation x + x 6 = 0 as te roots, 3. Te domain is Solutions to te take ome part (, 3) ( 3, ) (, ).. Explain wat it means tat two functions are equal. Tat is, complete te following (long) sentence. If f and g are functions, ten f = g if and only if : If f and g are functions, ten f = g if and only if tey ave te same domain and f(x) = g(x) for eac x in te domain. Based on your explanation, decide if te following functions f, g are equal or different. In eac case, justify your answer. (a) f(x) = x + x, g(t) = t + t. Equal; te domains are te same and for eac point of te domain tey assume te same value. (b) f(x) = x, x + g(x) = x. Not really equal. Te domain of te first one is (, ) (, ), of te second one (, ). (c) f(u) = sin u, g(x) = tan x. cos u Equal. Te tangent is by definition te sine over te cosine. (d) f(x) = e ln x, g(x) = x. Not equal. Te domain of te first one is (0, ), of te second one (, ). (e) f(x) = ln(e x ), g(x) = x. Equal.. Give examples of functions aving te following properties: (a) A function f of domain (, ) suc tat f(s + t) = f(s) + f(t). For example, f(x) = 0 for all x. More interesting, f(x) = x. (b) A function f of domain (, ) suc tat f(s + t) is NOT always equal to f(s) + f(t). Almost any function will work. For example, f(x) = x. (c) A function f of domain (, ) suc tat f(s + t) = f(s)f(t). Once again f(x) = 0 for all x is an example. More interesting, f(x) = x.

3 SOLUTIONS TO THE TAKE HOME PART 3 3. (a) Wat is a one-to-one function? How can you tell tat a function is one-to-one by looking at its grap? A one-to-one function is one tat assigns different values to different points in te domain. precisely: A function f of domain D is one-to-one if (and only if) f(s) f(t) wenever s, t D and s t. A completely equivalent, sometimes preferred, definition is: A function f of domain D is one-to-one if (and only if) f(s) = f(t) implies s = t. From a grap point of view, a function is one-to-one if and only if every orizontal line intersects te grap of f at most once. (b) If f is one to one, ow is its inverse function f obtained? How are te domain and range of f related to tose of f? Te inverse function can be obtained by solving for eac y in te range of f, te equation f(x) = y. Tis gives x as a function of y; tat is, it produces x = f (y). One can ten, if one wises, switc variables to get y = f (x). Te domain of f is te range of f, te range of f is te domain of f. (c) How do you obtain te grap of f from te grap of f? By reflecting te grap of f wit respect to te line y = x. 4. Te sine function y = sin x wen restricted to te interval [ π/, π/] is one-to-one; its inverse is te arc sine function, denoted by y = sin x or y = arcsin x. (a) Wat is te range and domain of y = arcsin x? Te range of arcsin is te domain of sin (i.e., te domain to wic sin was restricted; tat is [ π/, π/]. Te domain of arcsin is te range of sin, namely [, ]. (b) Simplify: tan(arcsin x). Tat is, find an expression tat does not involve any trigonometric functions. I give two solutions. Solution., te easy one, by pictures. Tink of arcsin x as an angle θ; tat is set t = arcsin x. Ten sin θ = x. We ave to find tan θ. Draw a rigt triangle, make on angle equal to θ and in te simplest way possible give values to a leg and ypotenuse so tat sin θ works out to x. Use te teorem of Pytagoras to find te value of te remaining leg. Te simplest way is to say te ypotenuse equals, te leg opposite to te angle is x. Ten Pytagoras gives tat te oter leg is x. From te picture we can see tat tan θ = x x. Solution., no pictures. Set again θ = arcsin x so sin x = θ. Now sin θ + cos θ = so tat cos θ = ± sin θ = ± x. We ave to decide if te sign is + or. Te range of arcsin is [ π/, π/]; in tat interval cos is non-negative, so tat cos θ = + x. tus 5. Find te domain and range of te following functions. tan(arcsin x) = tan θ = sin θ cos θ = x. x (a)f(x) = x 4 (b) g(x) = 64 x 4 (c)(x) = ln(x + 6) x 9 (d) k(x) = sin x

4 SOLUTIONS TO THE TAKE HOME PART 4 In every case to find te range it may pay to try for a more or less primitive grap. I just write out te answers. You can see me in my office for details. (a) Te domain is (, ) (, ) (, ). Te range is (, ) (0, ). (b) Te domain is [, ]. Te range is (0, ). (c) Te domain is ( 6, 3) ( 3, 3) (3, ). Te range is (, ). (d) Te domain is {x (, ) : x 0, ±π, ±π, ±3π,...}. Te range is (, ] [, ). 6. Te following picture is te grap of a function. Te grap consists of te lower alf of te circle of radius centered at te origin, and te upper alf of te circle of radius 3 centered at te point of coordinates (4, 0). Express tis function (call it f, for example) as a piecewise defined function. f(x) = { x, if x, 9 (x 4), if < x Consider te function f(x) = x + x +. (a) Determine te domain and te range of f. (b) Determine te inverse function of f. Te domain is {x (, ) : x /}. Te range is a bit arder to determine, but if we do part (b) it will work itself out on its own. To find te inverse we solve for x a bit of algebra gives y = x + x + ; x = y y. Tus, it seems tat f (x) = x. We found te inverse. We also found te range of f; since te domain x of te inverse is {x (, ) : x /}, tat s te range of f. 8. If f(x) = and g(x) =, find an expression for te function f g and determine its domain. 6 x x Te domain of f g is te set of all points in te domain of g wic g sends to te domain of f. To be in te domain of g, we must ave x 0; ten g sends suc an x to /x; to be in te domain of f

5 SOLUTIONS TO THE TAKE HOME PART 5 we need /x < 6. Tis works out to x > 4 or x < 4. If x > 4 or if x < 4, it is certainly not 0, so te domain of f g is (, 4) (4, ). On tat domain, f g(x) = x. 9. Two functions f, g are defined by f(x) = {, if x < 0, x, if x 0. g(x) = { 3x, if x, 3, if x > Express g f as a piecewise defined function. If we talk tings out,it isn t too ard. If x < 0, ten f(x) =. Since, g( ) = 3( ) = 3. Tus g f(x) = 3 if x < 0. If x > 0, ten f(x) = x. Wat g now does depends on weter x or x. Suppose x. Ten g(x ) = 3(x ) = 3x 4. Notice tat since x 0, te condition x is te same as x. So we ave g f(x) = 3x 4 if 0 x. If x > ; equivalently if x >, ten g(x ) = 3. Tus g f(x) = 3 if x >. Putting it all togeter, 3, x < 0, g f(x) = 3x 4, 0 x, 3, x >.