1. Consider the trigonometric function f(t) whose graph is shown below. Write down a possible formula for f(t).

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1 . Consider te trigonometric function f(t) wose grap is sown below. Write down a possible formula for f(t). Tis function appears to be an odd, periodic function tat as been sifted upwards, so we will use sin(t) as our parent function and apply te appropriate transformations. Te grap appears ( ) to be a function wose amplitude is, period is 4π, and midline is. So t f(t) = sin +.. Find te domain of te function f(x) = 4 5x. Because te square root of a negative number doesn t produce real numbers, we will only allow our domain to be values of x suc tat 4 5x is not negative. In oter words, 4 5x 0. We solve tis inequality for x: 4 5x 0 5x 4 x 4 5 So te domain is (, 4/5]. 3. Wat is te domain and range of te function f(x) = (x + 3) 4? Tis function is a polynomial, so its domain is all real numbers. Te sape of te grap is a parabola tat opens upwards, so to find te range we first find te vertex. Te equation of f(x) is already in vertex form, so we know te vertex is ( 3, 4). Since te grap as a minimum at its vertex and extends upwards infinitely, te range is [ 4, ).

2 4. Use te it definition of derivative to compute te derivative of te function f(x) = 4 x at x =. Te it definition of a derivative is f f(x + ) f(x) (x). Using tis definition, f (x) f(x + ) f(x) 4 x+ 4 x 4 x+ ( x x ) 4 x ( x+ x+ ) [Since f(x) = 4 x ] [Common denominators] 4x x(x+) 4(x+) x(x+) 4x 4(x+) x(x+) 4x 4x 4 x(x+) 4 x(x+) 4 x(x + ) ( ) [Distributive property] [Dividing by is te same as multiplying by ] 4 x(x + ) 4 x(x + ) = 4 x(x + 0) [Cancellation of ] [Evaluate it by setting = 0] = 4 x [Evaluate it by setting = 0] Te derivative at x = is f () = 4 ( ) =.

3 5. Te table below gives te dept of snow tat as fallen in inces as a function of time in ours past 8am. Wat is d(.5) (including units) and wat does it represent? Wat time did it start snowing? Next, estimate d () and d (). Include units in your answer and say in a full Englis sentence wat te meaning of tese numbers are. x d(x) d(.5) =.5 inces, wic represents te dept of te snow at.5 ours past 8 am, (wic is 9:30 am). Since d(.75) = 0 inces, and after tat time te dept of snow is positive, we can conclude tat it started snowing between 8:45 am and 9:00 am. We can estimate d () by calculating te slope between x = and x =.75: d() d(.75).75 = 0.3 =. inces/r. 0.5 We can estimate d () in te same way, by calculating te slope between x = and x =.75: d() d(.75).75 = = 0.6 inces/r. Te interpretation is tat d (). inces/r means tat at 9 am, te dept of te snow was increasing at. inces per our. Likewise, d () 0.6 inces/our means at 0 am, te dept of te snow was increasing at.06 inces per our. 6. Say f(x) = 3x + x. Use te definition of derivative to find f (). Ten write te equation of te tangent line to te curve at x =. f f( + ) f() 3( + ) + ( + ) (3() + ) () = 7. f () = 7 is te slope of te tangent line at x =, and f() = 4, so te equation of te tangent line to te curve is y 4 = 7(x ). 3

4 7. Say tat te position of an object moving orizontally is given by s(t) = 6t + 7, were position is measured in miles and time is measured in ours. Find te instantaneous velocity at an arbitrary t = a, and ten te instantaneous velocity at t = 7. Include units. Te instantaneous velocity at a time t is given by s (t), s s(a + ) s(a) 6(a + ) + 7 6a + 7 (a) ( ) 6(a + ) + 7 6a + 7 6(a + ) a + 7 6(a + ) a + 7 s 3 (7) = = 3 6(7)+7 7 6(a + ) + 7 (6a + 7) ( 6(a + ) a + 7) 6 ( 6(a + ) a + 7) 6 ( 6(a + ) a + 7) 6 6a =. [Tis is instantaneous velocity at t = a] 6a + 7 miles/our. [Tis is instantaneous velocity at t = 7] 8. Consider te function x + cos x wen 5 < x < 0 f(x) = 5 wen x = 0 e 3x/ wen 0 < x < 5 (a) Define wat it means for a function f(x) to be continuous at a point x = a. A function f(x) is continuous at a point x = a wen f(x) exists and it x a is equal to f(a). (b) (c) (d) f(x) = x + f(x) = x 0 + f(x) = x 0 (e) x 0 f(x) = x e3x/ = e 3. + x 0 e3x/ = e 0 =. + x 0 x + cos x = 0 + cos 0 =. x 0 x 0 f(x) f(x) =. + (f) Is f(x) continuous at x = 0? Fully explain your answer. If it is not continuous, state wat type of discontinuity it is and wy. Te it as x approaces zero exists (we calculated it in part (e)), but it is not equal to f(0). Terefore, f(x) is not continuous at x = 0. Note tat te discontinuity at x = 0 is a removable discontinuity because te f(x) exists. x 0 4

5 9. Sketc te grap of a function g(x) tat satisfies ALL of te following properties: (a) g(x) = x + (b) g(x) does not exist. x (c) (d) (e) g(x) = 4 g(x) = x 4 + g(x) = x

6 0. Find te average velocity over te interval 0. t 0.3 of a car wose position s(t) is given by te following table. Ten estimate te velocity at t = 0.3. Include units. t (sec) s(t) (ft) Te average velocity between t = 0. and t = 0.3 is s(0.3) s(0.) = = 5 ft/sec. To estimate te velocity at t = 0.3 better, let us also calculate te average velocity between t = 0.3 and t = 0.4 and ten average te two average velocities. Te average velocity between t = 0.3 and t = 0.4 is s(0.4) s(0.3) =.8 0. = 8 ft/sec. Ten taking te average, = 6.5 ft/sec, so 6.5 feet per second is an estimate for te velocity at t =

7 . Te function f(t) = 6t + 64t gives te distance above te ground of a ball tat is trown from ground level straigt up into te air at time t = 0, wit an initial velocity of 64 ft/sec. (a) How ig is te ball above te ground at t = second? f() = 6() + 64() = 48 ft. (b) How fast is te ball moving at t =, t =, t = 3 and t = 4 seconds? First we calculate te derivative, f (t) by using te it definition. After we do tis, we sould get tat f (t) = 3t To calculate te velocity at t =, t =, t = 3, and t = 4, we substitute tese values into f (t). f () = = 3 ft/sec f () = 3() + 64 = 0 ft/sec f (3) = 3(3) + 64 = 3 ft/sec f (4) = 3(4) + 64 = 64 ft/sec. (c) Wat is te maximum eigt of te ball? (Hint: you sould ave gotten a velocity of 0 for one of te values of t in te last part) Since te ball travels upward and ten starts to fall, we see tat te maximum occurs wen te ball as zero velocity at te peak of its trajectory. At time t =, te velocity is zero, so tis must be te point at wic te ball stops traveling upward (i.e. positive velocity) and begins its traveling downward (i.e. negative velocity). Te maximum eigt is te eigt at t =, so f() = 6() + 64() = 64 ft.. Precisely state te Intermediate Value Teorem. A polynomial p(x) as p( ) = 3 and p(0) = and p() =. Sow tat p(x) as at least two zeroes. IVT: If f(x) is continuous on a closed interval [a, b] and N is any number between f(a) and f(b) (were f(a) f(b)), ten tere is a number c in (a, b) suc tat f(c) = N. Since p(x) is polynomial, it is continuous everywere. Tus we may use te IVT. p( ) = 3 and p(0) =. Since 0 is between 3 and -, te IVT states tat tere must exist a value c between - and 0 suc tat p(c ) = 0. So p(x) as a zero between and 0. Similarly, we know p(0) = and p() =. Again, 0 is between - and, so tere must exist a value c between 0 and suc tat p(c ) = 0. So p(x) as a zero between 0 and. Tus p(x) as at least two zeros. 7

8 3. Find te following its. Sow all of your work. x + 4x (a) x x x + 4x (x )(x + 6) x x x (x ) x (x + 6) = 8 3 x (b) x 9 x 9 3 x x 9 x 9 x 9 x 9 x 9 (3 x) (3 + x) (x 9) (3 + x) 9 x (x 9)(3 + x) 3 + x = 6 x (c) x 4 x x 4 x = 8 ( x ) (x) (x 4) (x) 4 x (x 4)(x) 8

9 (d) 3x 4 + x 4 x 4 3x 4 + x 4 x 4 (3x 4 + x ) ( ) x 4 (4 x 4 ) ( ) x x 3 x 4 4 x 4 = = 3 (e) 3x + x 4 x 3x + x 4 x (3x + x ) ( x ) (4 x) ( x ) 3x + x 4 x = (f) 4x + x x + 4x + x x + 4x + x ( x ) (x + ) ( x ) 4x + x + x 4 + x x + x = + 0 = x 9

10 (g) ( 9x + x 3x) x ( 9x + x 3x) = + x = (Warning: Wile it is true tat + =, it is not true tat = 0, Observe te following problem.) () ( 9x + x 3x) ( 9x + x 3x) ( 9x + x 3x) ( 9x + x + 3x) ( 9x + x + 3x) (9x + x) 9x 9x + x + 3x = x 9x + x + 3x x ( ( x ) 9x + x + 3x) ( x ) 9 + x + 3 = 6 0

11 4. Te curve below sows position s, measured in feet, as a function of time t, measured in seconds. Te dotted line is tangent to te curve. 6 s(t) t (a) (b) Find te average velocity over te interval from t = 0 to t = 3. Include units. s(3) s(0) average velocity = = 0 5 = ft/s Find te instantaneous velocity at t =. Include units. Te instantaneous velocity is te slope of te tangent line at t =. Tis line goes troug (, 5) and (, 3). So te instantaneous velocity is s () = 3 5 = ft/sec. (c) Find te equation of te tangent line. Te equation of a line troug (, 3) wit slope - is s 3 = (t ) or s = t + 7.

12 5. Here s a grap of te function f(x): Meanwile, ere is a definition of te function g(x): (x 3) + x > 3 g(x) = x + 6 x < 3 (a) Explain wy te grap defines f as a function of x. It passes te vertical line test. (b) Is te f function invertible? No, it fails te orizontal line test. (c) Is te function g invertible? By sketcing g, we can see tat it is inveritble (d) Wat is te domain and te range of f? Te domain of f is ( 4, ) (, 4) (4, 5) and te range of f is ( 3, ). (e) Wat is te domain and te range of g? Te domain of g is (, 3) (3, ) and te range of g is (, ) (3, ).

13 (f) Find te it of f(x) as x approaces -, as x approaces, as x approaces, as x approaces 3, as x approaces 4. Clearly explain your results. x f(x) = DNE; x + f(x) = 3 but f(x) = ; removable discontinuity x f(x) = ; f is continuous at x x 3 f(x) = DNE; f(x) = 4 but x 3 + f(x) = 4; removable discontinuity f(x) = x x 3 f(x) = (g) Wat is te it of g(x) as x approaces 3? Wat about as x approaces 0? x 3 x 0 g(x) = DNE; g(x) = but x 3 + g(x) = 6; g is continuous at 0 g(x) = 3 x 3 () List all of te discontinuities of f and g and clearly explain wy tese are discontinuities. State wat type of discontinuities tey are and wy. Discontinuities of f: x = 4 : f(x) exists, but f( 4) is undefined, a removable discontinuity x 4 x = : f(x) =, an infinite discontinuity x x = : f(x) = but f() =, a removable discontinuity x x = 3 : f(x) f(x), a jump discontinuity since bot sided its exist x 3 x 3 + x = 4 : f(x) = 4 but f(4) is undefined, a removable discontinuity x = 5 : f(x) exists, but f(5) is undefined, a removable discontinuity x 5 Discontinuities of g: x = 3 : g(x) g(x), a jump discontinuity since bot sided its exist x 3 x 3 + 3

14 (i) (f(x) + g(x)) = x (f(x) + g(x)) f(x) + g(x) = + 4 = 5 x x x (j) x 3 (f(x) + g(x)) = x 3 x 3 x 3 + g(x)) f(x) + +(f(x) g(x) = + 4 = 5 x 3 + x g(x)) f(x) + (f(x) g(x) = + 3 = 5 x 3 x 3 (f(x) + g(x)) = 5 6. Find two values of te constant b so tat te following function is continuous: x + x x b (x) = 3x + x < b For (x) to be continuous, we need (x) (x). We ave (x) = x b + x b x b + b + b and (x) = 3b +. Setting tem equal gives, x b b + b = 3b + b + 4b = 0 (b + 6)(b ) = 0 b = 6 or b =. 7. If 4x 9 f(x) x 4x + 7 for x 0, ten find f(x). By te Squeeze Teorem, Tus, f(x) = 7. 4x 9 f(x) x 4x + 7 4(4) 9 f(x) 4 4(4) f(x) Suppose f() = 3 and f () =. Find f( ) and f ( ) if f is assumed to be even. f( ) = 3 and f ( ) =. 4

15 9. Given all of te following information about a function f, sketc its grap. f(x) = 0 at x = 5, x = 0, and x = 5 f(x) = x f(x) = 3 f (x) = 0 at x = 3, x =.5, and x = 7 Tere are many viable functions tat can satisfy tese conditions. Here is a roug sketc of one possible function: 5