Math 19 Sample Final Exam Solutions

Transcription

1 SSEA Summer 2017 Math 19 Sample Final Exam Solutions 1. [10 points] For what value of the constant k will the function: ) 1 x 2 sin, x < 0 fx) = x, x + k coskx), x 0 be continuous at x = 0? Explain your answer clearly. f is continuous at x = 0 if: lim fx) = lim fx) = f0) x 0 x 0 + ) 1 lim x2 sin = 0 + k cosk 0) x 0 x ) 1 lim x2 sin = k. x 0 x ) 1 We can evaluate lim x 0 x 2 sin using the squeeze theorem as follows: x ) 1 1 sin 1 x 2 ) 1 x 2 x 2 sin x 2 x 2 ) 1 lim x2 lim x2 sin lim x 0 x 0 x 2 x2 x 0 ) 1 0 lim x2 sin 0 x 0 x ) 2 1 lim x2 sin = 0. x 0 x 2 Therefore, k = 0 is require for f to be continuous at x = 0.

2 SSEA Math 19 Sample Final Exam Solutions, Page 2 of 6 August 2, [7 points] Is there a value of b that will make gx) = x = 0? Differentiable at x = 0? Explain your answers. { x + b, x < 0 cosx), x 0 continuous at f is continuous at x = 0 if: lim x 0 x 0 fx) = lim fx) = f0) + lim x 0 x + b = cos0) 0 + b = 1 b = 1. Next, we want f to be ifferentiable at x = 0 x + b) = x=0 cosx)) x=0 1 = sinx) x=0 1 = 0, which is not possible. Therefore, f is not ifferentiable at x = 0 for any real value of b.. [8 points] Suppose that functions f an g an their erivatives with respect to x have the following values at x = 2: fx) gx) f x) g x) π/2 2 1/ - Compute the erivative of [sin fx))] 2 + gx) ) 1/ at x = 2. [sin fx))] 2 + gx) ) 1/ 1 = [sin fx))] 2 + gx) ) 1 1 [sin fx))] 2 + gx) ) [ 2/ = 1 = 1 [sin fx))] 2 + gx) ) 2/ [sin fx))] 2 + gx) ) [sinfx))] 2 ) + ] gx)) [ [2 sinfx))] 2 1 +g x)] = 1 [sin fx))] 2 + gx) ) 2/ [ ] 2 sinfx)) cosfx)) fx)) + g x) sinfx)))

3 SSEA Math 19 Sample Final Exam Solutions, Page of 6 August 2, 2017 Evaluating at x = 2 = 1 [sin fx))] 2 + gx) ) 2/ [2 sinfx) cosfx))f x) + g x)]. [sin fx))] 2 + gx) ) 1/ 1 = [sin f2))] 2 + g2) ) 2/ [2 sinf2)) cosf2))f 2) + g 2)] = 1 [sin π/2)] ) 2/ [2 sinπ/2) cosπ/2)1/) ] = ) 2/ [ /) ] = [8 points] Fin equations for the tangent an normal to the curve y 2 2 x) = x at the point 1, 1). Hint: Implicit ifferentiation) y 2 2 x) ) = x ) 2y y ) 2 x) + y 2 1) = x 2 y = x2 + y 2 2y2 x)) y = x2 + y 2 1,1) 2y2 x)) y = ,1) )) = 2, which is the slope of the tangent at 1, 1). So, the slope of the normal at 1, 1) is 1/2. The equation of the tangent is y 1 = 2x 1) an that of the normal is y 1 = 1 x 1). 2 1,1) 5. [5 points] Use an appropriate linear approximation to fin an approximation of the function fx) = 4 + x) 1/ for values of x near 0. 4+x) 1/ = )) 1/ x = 4 1 1/ + 4 ) 1/ x 4 1/ )) 4 x = 4 1/ 1 + x ). 4

4 SSEA Math 19 Sample Final Exam Solutions, Page 4 of 6 August 2, [6 points] Let f be a ifferentiable function for all values of x. Suppose the following conitions hol: f1) = 1, f < 0 on, 1), f > 0 on 1, ). Then, show that fx) 1 for all x. Hint: Mean Value Theorem) Since f exists on, 1) an 1, ), it is also continuous in those intervals. Applying the mean value theorem on the interval x, 1), where x < 1 f1) fx) = f c), where x < c < 1 1 x f1) fx) = ve number +ve number f1) fx) = ve number. This shows that on the interval, 1), we have fx) > f1) fx) > 1. Similarly, applying the mean value theorem on the interval 1, x), where x > 1 gives again that fx) > f1) fx) > 1. Putting these results with the fact that f1) = 1 implies fx) 1 for all x. 7. [8 points] Consier the graph of the first erivative of a function f shown below:

5 SSEA Math 19 Sample Final Exam Solutions, Page 5 of 6 August 2, 2017 a) Estimate the intervals on which f is increasing. f increases in the interval where f x) > 0. This happens in [, 2] an [1, 2]. b) Estimate the intervals on which f is ecreasing. f ecreases in the interval where f x) < 0. This happens in [ 2, 0) an 0, 1]. c) Use the given graph of f to inicate where any local extreme values of the function f occur, an whether each extreme is a local maximum or minimum. Provie that f is continuous at x = 0, local maximum values occur at x = 2 an at x = 2; local minimum values occur at x =, an at x = [8 points] What value of a makes fx) = x 2 + a/x) have: a) a local minimum at x = 2? This happens if f 2) = 0 an f 2) > 0: f 2) = 0 x 2 + a x) = 0 x=2 2x a x 2 ) x=2 = a 2 2 = 0 a = 16. For this value of a, the secon erivative f x) = 2 + 2a/x ) is greater than zero at x = 2 thus confirming that f has a local minimum at x = 2. b) a point of inflection at x = 1? f has an inflection point at x = 1 if: f 1) = a ) = x x=1 2a ) 1 = 0 a = [10 points] 10. [10 points] The 800-room Mega Motel chain is fille to capacity when the room charge is $50 per night. For each $10 increase in room charge, 40 fewer rooms are fille each night. There is a clean up cost of $5 per room per night. What charge per room will result in the maximum revenue per night?

6 SSEA Math 19 Sample Final Exam Solutions, Page 6 of 6 August 2, 2017 Let x be the number of $10-increases. For each $10 increase, 40 fewer rooms are fille; therefore, for x $10-increases, 40x fewer rooms will be fille, or the total number of rooms fille will be x. The rate per night for each room now is $ x because we a x $10- increases to the base rate of $50. So, the total revenue per night from renting the rooms is: x) }{{} x). }{{} # of rooms rente charge per room The clean up cost for each room per night is $800 40x) 5. The total revenue is rx) = x) x) x) 5, which we want to maximize. rx) = 400x x We compute r x) an set it equal to zero. r x) = 0 800x = 0 x = 6200/ So, x = 7.8 is a critical point of rx). We calculate r x) an evaluate it at x = 7.8 to etermine the nature of that critical point. r x) = 800, which is lesser than zero. So, x = 7.8 maximizes rx). Since x the # of $10-increases) is an integer, we choose x = 8. Therefore, the optimal room rate per night is $ x = $ = $10.