Sample Final Exam Problems Solutions Math 107

Transcription

1 Sample Final Eam Problems Solutions Math (a) We first factor the numerator and the denominator of the function to obtain f() = (3 + 1)( 4) 4( 1) i To locate vertical asymptotes, we eamine all locations where the denominator is equal to zero We have 4( 1) = 0 = = 0 or = 1 A graph or table confirms that f has vertical asymptotes at = 0 and at = 1 ii To locate horizontal asymptotes, we note that the degree of the numerator and the degree of the denominator are the same; they both equal 2 Therefore, since 3 is the leading coefficient of the polynomial in the numerator, and since 4 is the leading coefficient of the polynomial in the denominator, we get a horizontal asymptote of y = 3 4 iii To find -intercepts, we set y = 0 and solve for : (3 + 1)( 4) 4( 1) = 0 = (3 + 1)( 4) = 0 = = 1 3 or = 4 (b) Therefore, there are two -intercepts: 1 3 and 4 iv To find the y-intercept, we set = 0 and solve for y; that is, we calculate f(0) But f(0) is undefined because the denominator of f equals zero when = 0 Therefore, the function f has no y-intercept i To locate vertical asymptotes, we eamine all locations where the denominator is equal to zero However, since the equation = 0 has no real solutions, it follows that g has no vertical asymptotes ii To locate horizontal asymptotes, we note that the degree of the numerator is 1, and the degree of the denominator is 2 Therefore, g has a horizontal asymptote at y = 0 iii To find -intercepts, we set y = 0 and solve for : = 0 = + 1 = 0 = = Therefore, 1 is the only -intercept of g iv To find the y-intercept, we set = 0 and solve for y; that is, we calculate g(0) We have g(0) = = 1 3, so 1 3 is the y-intercept 2 Using the law of cosines, we have c c 2 = (3)(4)cos(15 ) = c 2 = 1818 = c = 1348 (a) The distance between the planes is therefore c = 1348 miles (b) Using facts about right triangles, we have 3 O O h sin(25 ) = h 4 = h = 1690 miles 3 (a) The function h is linear because the inputs are equally spaced ( = 1) and the outputs are also equally spaced ( y = 04) The function k is eponential because ratios of successive output values are equal 16 (ie 64 = 4 16 = 1 4 = = = 025) The function g has to be the trigonometric function because it increases, then decreases, and then increases again, and none of the other three types of functions can ehibit this behavior This means that f must be the power function by the process of elimination

2 Math 107 Sample Final Eam Problems Solutions 2 (b) Since k is eponential, we know that k() = ab, and we are given that k(1) = 64 and k(2) = 16 We must solve for a and b We have ab 1 = 64 = ab 2 = 16 ab 1 = 64 = a ab 2 ab 1 = = b = 1 4 ( ) 1 1 = 64 = a = 256, 4 and so our final answer is k() = 256 ( 1 4) (c) Since h is linear, we know that h() = m + b To find b, we note that h(1) = 134, so we have m = slope = y = = h() = 04 + b = 134 = 04(1) + b = b = 174, and so our final answer is h() = (d) Since f is a power function, we know that f() = k p, and we are given that f(1) = 05 and f(2) = 141 k 1 p = 05 = k 2 p = 141 k 2 p k 1 p = and so our final answer is f() = = 2 p = 282 = p = ln(282) ln 2 k 1 p = 05 = k = 05 = k = 05, (e) Since g is a trigonometric function, we know that it must have the form g() = A sin(b( h)) + k or g() = Acos(B( h)) + k, depending on what point we choose to be our starting point To help us decide, we first plot the points on the function g below: If we let the point (2, 05) be our starting point, then the function starts from its midline and increases, just like the sine function; therefore, we will use the formula g() = Asin(B( h)) + k to model the function g Furthermore, since we have chosen 2 as our starting point, it follows that our horizontal shift is given by h = 2 Now, from the plot, we can see that the midline is given by k = 05, and since the distance from the midline to the maimum value of the function is = 025, we see that the amplitude is given by A = 025 To finish, we must find the period and use this to figure out the constant B From our graph, we see that the pattern completes one full cycle between 1 and 5, meaning that the period is 5 1 = 4 Therefore, we have 2π B = 4 = B = π 2, and so our final answer is ( π ) g() = 025 sin 2 ( 2) + 05 We note that there are many other correct answers besides the one shown above

3 Math 107 Sample Final Eam Problems Solutions 3 4 (a) 3 t = 2e t 3 = ln(3 t ) = ln(2e t 3 ) = t ln 3 = ln 2 + ln e t 3 = t ln 3 = ln 2 + t 3 = t ln 3 t = ln 2 3 = t(ln 3 1) = ln 2 3 (b) Therefore, our answer is t = ln 2 3 ln 3 1 lnp = ln 5 + ln e kt = lnp = ln 5 + kt = lnp ln 5 = kt (c) Therefore, our answer is t = lnp ln 5 k 5 = 6 t 3 6 t + 2 = 6(t + 2) 6(t 3) 5 = (t 3)(t + 2) (t 3)(t + 2) = 6(t + 2) 6(t 3) 5 = (t 3)(t + 2) = 5(t 2 t 6) = 6t t + 18 = 5t 2 5t 30 = 30 = 5(t 2 t 12) = 0 = 5(t 4)(t + 3) = 0 Therefore, there are two solutions: t = 4 and t = 3 (d) Using the identity sin 2 t + cos 2 t = 1, we note that cos 2 t = 1 sin 2 t is also an identity Therefore, we can substitute 1 sin 2 t for cos 2 t to rewrite the given equation as 2(1 sin 2 t) = 3 sin t +3 We therefore have 2(1 sin 2 t) = 3 sint + 3 = 2 2 sin 2 t = 3 sint + 3 = 2 sin 2 t + 3 sint + 1 = 0 = (2 sin t + 1)(sint + 1) = 0 = sin t = 1 2 or sin t = 1 Thus, there are two possibilities we need to consider: either sint = 1 or sint = 1 2 Possibility 1: Let us first consider the possibility that sint = 1 In this case, we are looking for all angles t that have a sine of 1 Since t = 3π 2 is the only angle on the unit circle that yields a y-coordinate of 1, we see that t 1 = 3π 2 is the only solution to this part of the problem Possibility 2: Now, we consider the possibility that sint = 1 2 We first calculate a reference angle: ( ) 1 t = sin 1 = π 2 6 Since the sine function is negative in quadrants III and IV, we have one solution from each quadrant They are: t 2 t 3 t 2 = π + π 6 = 7π 6 t 3 = 2π π 6 = 11π 6 Therefore, there are three solutions to the original problem: t = 3π 2, t = 7π 6 and t = 11π 6

4 Math 107 Sample Final Eam Problems Solutions 4 5 (a) i This represents the air pressure, in pounds per square inch, at an elevation of 20 thousand feet above sea level ii This represents the elevation, in thousands of feet, at which the air pressure is 20 pounds per square inch iii This represents the height, in feet, of the ball 4 seconds after it has been thrown upward (b) No, it does not have an inverse function because the ball falls to the ground after being thrown upward Therefore, the graph of the height of the ball as a function of time fails the horizontal line test 6 (a) i We have f( 1) = 2 ( 1) = 3 1 ii Calculating, we see that f(g()) = f ( ) = = 2(2 ) 2 2 = = (b) No, g() is not the inverse of f() because f(g()) (see part (ii) of (a) from above) Another way to see that g() is not the inverse of f() is to actually calculate the inverse of f() and see for ourselves that it does not equal g(), as demonstrated below: y = 2 = y = 2 = y + = 2 = (1 + y) = 2 = = y As a result, we see that f 1 (y) = 2 1+y, and therefore that f 1 () = Since g() 1+, we again see that g() is not the inverse function of f() 7 Let us start by assuming that t = 0 represents the year 1960 (a) Know: P = ab t = b t Given: P = when t = 30 Since P = when t = 30, we have P = b t = = b 30 = 4 = b 30 = b = 4 1/ Therefore, our final answer is P = 10000(10473) t (b) In the year 1970, we have t = 10, so our answer is P = 10000(10473) people Therefore, the town reaches this population sometime in 1983 (c) We have P = = 10000(10473) t = = t ln(10473) = ln 3 (d) The average rate of change is given by P t = = t = 2377 years = 5875 people per year 8 (a) We have f(1 + 5) = f(6) 265 feet per second, which is the speed of the car 6 seconds after the brakes are applied

5 Math 107 Sample Final Eam Problems Solutions 5 (b) We have f(1) = 95 feet per second, which represents a speed 5 miles per hour faster than the speed of the car 1 second after the brakes are applied (c) From the graph, we can see that we achieve an output value of 40 when the input value is approimately 47 t , so solving the previous equation for t, we obtain t 47 2 = 27 Thus, our answer is t 27 seconds (d) We have therefore, t 57 seconds is our answer f(t) + 10 = 40 = f(t) = 30; 9 Before beginning on the individual parts of the problem, note that from the form of the equations, we can see that y = m + b is a linear function, y = k p is a power function, y = qr is an eponential function, and y = a cos + c is a trigonometric function (a) The constant b is definitely negative since the line shown in the graph intersects the negative portion of the y-ais, meaning that the y-intercept is negative The constant c is also definitely negative because it represents the midline of the trigonometric function that is shown, and we can see that the midline of this function lies below the -ais (b) The constant k is definitely positive because the graph of the power function y = k p lies above the -ais Similary, the constant q is definitely positive because the eponential function shown lies above the -ais The constant m is definitely positive because it represents the slope of the line that is shown, and we can see that the line rises from left to right The constant p is definitely positive because the graph of the power function is increasing Finally, the constant r is definitely positive because eponential functions must always have positive bases, by definition (c) The constant p is definitely greater than zero but less than one because the power function is concave downward (like the graph of y = 1/2, for eample) The constant r is definitely greater than zero but less than one because the eponential function y = qr is decreasing, implying that its base must be less than one (d) We cannot say for sure than any of the constants are greater than one Therefore, none of the constants are definitely greater than one