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NAME: Calculus with Analytic Geometry I Exam 1, Friday, August 30, 2011 SOLUTIONS 1 Solutions to the in class part 1. The graph of a function f is given. (a) State the value of f (1). (b) Estimate the value of f (−1). (c) For what values of x is f (x) = 1? (d) Estimate the value of x such that f (x) = 0. (e) State the domain and range of f . (f) On what interval is f increasing. Solution. (a) f (1) = 3. (b) f (−1) ≈ −1/3 (any value in the range −0.4 to −0.2 is acceptable.) (c) For x = 1 and x = 3. (d) f (−1) ≈ −2/3 (any value in the range −0.75 to −0.6 is acceptable.) (e) The domain is (−2, 4). The range is (−1, 2]. (f) On the interval (−2, 1]. f (a + h) − f (a) 2. If f (x) = x3 , evaluate . Your final expression should be as simple as possible. That is, h anything that can be canceled or simplified in any way, should be canceled and simplified. 2 SOLUTIONS TO THE TAKE HOME PART 2 Solution. f (a + h) − f (a) h = = = (a + h)3 − a3 a3 + 3a2 h + 3ah2 + h3 − a3 6 a3 + 3a2 h + 3ah2 + h3 − 6 a3 = = h h h 3a2 h + 3ah2 + h3 (3a2 + 3ah + h2 )h (3a2 + 3ah + h2 ) 6 h = = h h 6h 3a2 + 3ah + h2 . 2x3 − 5 . x2 + x − 6 Solution. The only problem this function has is that the denominator can be 0. The quadratic equation x2 + x − 6 = 0 has the roots 2, −3. The domain is 3. Find the domain of the function f (x) = (−∞, −3) ∪ (−3, 2) ∪ (2, ∞). 2 Solutions to the take home part 1. Explain what it means that two functions are equal. That is, complete the following (long) sentence. If f and g are functions, then f = g if and only if Solution. : If f and g are functions, then f = g if and only if they have the same domain and f (x) = g(x) for each x in the domain. Based on your explanation, decide if the following functions f, g are equal or different. In each case, justify your answer. (a) f (x) = x2 + 1 1 , g(t) = t2 + . 1−x 1−t Equal; the domains are the same and for each point of the domain they assume the same Solution. value. x2 − 1 (b) f (x) = , g(x) = x − 1. x+1 Solution. Not really equal. The domain of the first one is (−∞, 1) ∪ (1, ∞), of the second one (−∞, ∞). sin u , g(x) = tan x. (c) f (u) = cos u Solution. Equal. The tangent is by definition the sine over the cosine. (d) f (x) = eln x , g(x) = x. Solution. Not equal. The domain of the first one is (0, ∞), of the second one (−∞, ∞). (e) f (x) = ln(ex ), g(x) = x. Solution. Equal. 2. Give examples of functions having the following properties: (a) A function f of domain (−∞, ∞) such that f (s + t) = f (s) + f (t). Solution. For example, f (x) = 0 for all x. More interesting, f (x) = x. (b) A function f of domain (−∞, ∞) such that f (s + t) is NOT always equal to f (s) + f (t). Solution. Almost any function will work. For example, f (x) = x2 . (c) A function f of domain (−∞, ∞) such that f (s + t) = f (s)f (t). Solution. Once again f (x) = 0 for all x is an example. More interesting, f (x) = 2x . 2 SOLUTIONS TO THE TAKE HOME PART 3 3. (a) What is a one-to-one function? How can you tell that a function is one-to-one by looking at its graph? Solution. A one-to-one function is one that assigns different values to different points in the domain. precisely: A function f of domain D is one-to-one if (and only if) f (s) 6= f (t) whenever s, t ∈ D and s 6= 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 graph point of view, a function is one-to-one if and only if every horizontal line intersects the graph of f at most once. (b) If f is one to one, how is its inverse function f −1 obtained? How are the domain and range of f −1 related to those of f ? Solution. The inverse function can be obtained by solving for each y in the range of f , the equation f (x) = y. This gives x as a function of y; that is, it produces x = f −1 (y). One can then, if one wishes, switch variables to get y = f −1 (x). The domain of f −1 is the range of f , the range of f −1 is the domain of f . (c) How do you obtain the graph of f −1 from the graph of f ? Solution. By reflecting the graph of f with respect to the line y = x. 4. The sine function y = sin x when restricted to the interval [−π/2, π/2] is one-to-one; its inverse is the arc sine function, denoted by y = sin−1 x or y = arcsin x. (a) What is the range and domain of y = arcsin x? Solution. The range of arcsin is the domain of sin (i.e., the domain to which sin was restricted; that is [−π/2, π/2]. The domain of arcsin is the range of sin, namely [−1, 1]. (b) Simplify: tan(arcsin x). That is, find an expression that does not involve any trigonometric functions. Solution. I give two solutions. Solution 1., the easy one, by pictures. Think of arcsin x as an angle θ; that is set th = arcsin x. Then sin θ = x. We have to find tan θ. Draw a right triangle, make on angle equal to θ and in the simplest way possible give values to a leg and hypotenuse so that sin θ works out to x. Use the theorem of Pythagoras to find the value of the remaining leg. The simplest way is to say the √ hypotenuse equals 1, the leg opposite to the angle is x. Then Pythagoras gives that the other leg is 1 − x2 . From the picture we can see that tan θ = √ x . 1−x2 2 2 Solution 2., p no pictures. √Set again θ = arcsin x so sin x = θ. Now sin θ + cos θ = 1 so that cos2 θ = ± 1 − sin2 θ = ± 1 − x2 . We have to decide if the sign √ is + or −. The range of arcsin is [−π/2, π/2]; in that interval cos is non-negative, so that cos θ = + 1 − x2 . thus tan(arcsin x) = tan θ = sin θ x =√ . cos θ 1 − x2 5. Find the domain and range of the following functions. (a)f (x) = 2 x2 − 4 (b) g(x) = p 64 − x4 (c)h(x) = ln(x + 6) x2 − 9 (d) k(x) = 1 sin x 2 SOLUTIONS TO THE TAKE HOME PART 4 Solution. In every case to find the range it may pay to try for a more or less primitive graph. I just write out the answers. You can see me in my office for details. 1 (a) The domain is (−∞, −2) ∪ (−2, 2) ∪ (2, ∞). The range is (−∞, − ) ∪ (0, ∞). 2 √ √ (b) The domain is [−2 2, 2 2]. The range is (0, ∞). (c) The domain is (−6, −3) ∪ (−3, 3) ∪ (3, ∞). The range is (−∞, ∞). (d) The domain is {x ∈ (−∞, ∞) : x 6= 0, ±π, ±2π, ±3π, . . .}. The range is (−∞, −1] ∪ [1, ∞). 6. The following picture is the graph of a function. The graph consists of the lower half of the circle of radius 1 centered at the origin, and the upper half of the circle of radius 3 centered at the point of coordinates (4, 0). Express this function (call it f , for example) as a piecewise defined function. Solution. f (x) = 7. Consider the function f (x) = √ 2 − if −1 ≤ x ≤ 1, p 1−x , 9 − (x − 4)2 , if 1 < x ≤ 7. x+1 . 2x + 1 (a) Determine the domain and the range of f . (b) Determine the inverse function of f . Solution. The domain is {x ∈ (−∞, ∞) : x 6= −1/2}. The range is a bit harder to determine, but if we do part (b) it will work itself out on its own. To find the inverse we solve for x y= x+1 ; 2x + 1 x= 1−y . 2y − 1 a bit of algebra gives 1−x . We found the inverse. We also found the range of f ; since the domain 2x − 1 of the inverse is {x ∈ (−∞, ∞) : x 6= 1/2}, that’s the range of f . Thus, it seems that f −1 (x) = 1 1 and g(x) = , find an expression for the function f ◦ g and determine its domain. x 16 − x2 Solution. The domain of f ◦ g is the set of all points in the domain of g which g sends to the domain of f . To be in the domain of g, we must have x 6= 0; then g sends such an x to 1/x; to be in the domain of f 8. If f (x) = √ 2 SOLUTIONS TO THE TAKE HOME PART 5 we need 1/x2 < 16. This works out to x > 4 or x < −4. If x > 4 or if x < −4, it is certainly not 0, so the domain of f ◦ g is (−∞, −4) ∪ (4, ∞). On that domain, 1 f ◦ g(x) = q 1− 9. Two functions f, g are defined by −1, if x < 0, f (x) = x2 , if x ≥ 0. . 1 x2 g(x) = 3x2 , if x ≤ 1, 3, if x > 1 Express g ◦ f as a piecewise defined function. Solution. If we talk things out,it isn’t too hard. If x < 0, then f (x) = −1. Since −1 ≤ 1, g(−1) = 3(−1)2 = 3. Thus g ◦ f (x) = 3 if x < 0. If x > 0, then f (x) = x2 . What g now does depends on whether x2 ≤ 1 or x2 ≥ 1. Suppose x2 ≤ 1. Then g(x2 ) = 3(x2 )2 = 3x4 . Notice that since x ≥ 0, the condition x2 ≤ 1 is the same as x ≤ 1. So we have g ◦ f (x) = 3x4 if 0 ≤ x ≤ 1. If x2 > 1; equivalently if x > 1, then g(x2 ) = 3. Thus g ◦ f (x) = 3 if x > 1. Putting it all together, x < 0, 3, 3x4 , 0 ≤ x ≤ 1, g ◦ f (x) = 3, x > 1.