MATH 236 ELAC FALL 2017 CA 9 NAME: SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

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1 MATH 236 ELAC FALL 207 CA 9 NAME: SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. ) 27 p 3 27 p 3 ) 2) If 9 t 3 4t 9-2t = 3, find t. 2) Solve the equation. 3) 3( + 2x) = 243 3) 4) 2(7 + 3x) = 4 4) 5) (e/2x ) -3/4 e 2x - 5 5) Enter your answer exactly as e P(x) where P is a polynomial with any fractions in the form a in lowest terms. b 6) If (e x ) 2 e 2x e =, find x. 6) e2 Differentiate. 7) f(x) = ( + x 2 )e x Enter your answer exactly in the form (P(x))e a where P is a polynomial. 7) 8) y = 8xe x - 8e x 8) 9) y = 5e x 2ex + 9) 0) y = e 4x2 + x 0) ) (6e 2x - x) 3 ) 2) y = e 7x/2 2) 3) Find the values of x at which the function f(x) = e -2x + 2x has a possible relative maximum or minimum point. 3)

2 Solve the problem. 4) The sales in thousands of a new type of product are given by S(t) = 70-80e -0.9t, where t represents time in years. Find the rate of change of sales at the time when t = 7. 5) Suppose that the amount in grams of a radioactive substance present at time t (in years) is given by A(t) = 380e -0.32t. Find the rate of change of the quantity present at the time when t = 5. 4) 5) 6) If ln x = -3.6, write x using the exponential function. 6) 7) ln(x 2-2) + e ln(x 2-2) 7) 8) e x + 2 ln x Enter your answer exactly as a b e c. 8) Solve for x. 9) 2 - ln(x + 3) = ln 4 9) 20) 2 ln x + ln x = 3 20) 2) 2e x = 6 Enter your answer exactly as just a ± ln b. 2) 22) 2 ln(x + ) - ln x 2 = 8 Enter your answer exactly as just a e b. - c 22) Solve the equation for x. 23) e -x = 9 23) Solve the problem. 24) A company begins an advertising campaign in a certain city to market a new product. The percentage of the target market that buys the product is a function of the length of the advertising campaign. The company estimates this percentage as - e -0.03t where t = number of days of the campaign. The target market is estimated to be,000,000 people and the price per unit is $0.60. The cost of advertising is $3000 per day. Find the length of the advertising campaign that will result in the maximum profit. 24) 2

3 25) The demand function for a certain product is given by 25) D(p) = 600e -0.p, where p is price per unit. Recall that total revenue is given by R(p) = pd(p). At what price per unit p will the revenue be maximum? 26) At what value of x could the function f(x) = ln x + x x minimum? have a possible relative maximum or 26) Find the derivative of the function. 27) y = ln (6 + x2) 27) 28) y = ln (ln 2x) 28) 29) y = ln (9x3 - x2) 29) Differentiate. x ) ln x - 30) 3) (ln(x 2 + 2)) 3 3) 32) ln 3x 32) 33) x 3 ln x 33) 34) (x + 3ln x) 4 at x = Enter just an integer. 34) 35) + ln (2 - x) at x = Enter your answer as just a reduced fraction a b. 35) 36) ln + x2 2x + 5 at x = 36) Enter your answer as just a reduced fraction a b. 37) ln(e x + e -x ) at x = 0 Enter just an integer. 37) 38) Find the slope of the graph of y = ln(2x + 3) /2 at the point (3, ln 3). 38) 3

4 39) Find an equation of the tangent line to the graph of y = 2x + ln x at x =. 39) Find the x-value of all points where the function has relative extrema. Find the value(s) of any relative extrema. 40) f(x) = x4 5lnx 40) Solve the problem. 4) Assume the total revenue from the sale of x items is given by R(x) = 24 ln (x + ), while the total cost to produce x items is C(x) = x/4. Find the approximate number of items that should be manufactured so that profit, R(x) - C(x), is maximum. 4) 42) Suppose that the demand function for x units of a certain item is p = 00 + where p is the price per unit, in dollars. Find the marginal revenue. 80 ln(x + 5), x 42) 43) ln y - 3[2ln (y - 6) - ln (y + 6)] 43) Solve for x. 44) ln x + ln x 8 = 9 44) Provide an appropriate response. 45) A study comparing the sizes of two populations, x and y, shows that they can be related by the equation ln( + y) - k ln x = ln C, where k and C are constants. Solve the equation for y when k = 5 and C = 7. 45) Differentiate. xe 46) ln x x 2 + at x = 46) Enter just a reduced fraction of form a b. 47) f(t) = ln [(t 6-5)(t 5 + 3)] 47) 48) y = ln - x (x + 3) 5 48) Use logarithmic differentiation to differentiate. 49) (3x + ) 5 (2x - ) -2 (x + 3) 4 at x = Enter your answer exactly as just 3 4 a. 49) Use logarithmic differentiation to find dy/dx. 50) y = 24-x 50) 4

5 Answer Key Testname: MATH236CA9 ) 3 (-5p - ) 2) t = 4 3) 2 4) -3 5) e (-9/8)x + 5 6) ) (x 2 + 2x + )e x 8) 8xe x 9) 5ex (2ex + )2 0) 8xe 4x2 + ) 3(6e 2x - x) 2 (2e 2x - ) 2) 7 2 e7x/2 3) minimum at x = 0 4) 0. thousand per year 5) grams per year 6) e 3.6 7) ( + e)ln (x 2-2) 8) x 2 e x 9) x = 4 e2-3 20) e 3 2) - + ln 2 22) e 4-23) x = -ln 9 24) 60 days 25) $0 26) x = e 2x 27) x ) x ln 2x 29) 27x - 2 9x2 - x 30) x x - 5

6 Answer Key Testname: MATH236CA9 6x 3) x 2 (ln(x 2 + 2)) ) 2x ln 3x 33) 3x 2 ln x + x 2 34) 6 35) ) ) 0 38) 9 39) y = x + 40) Relative minimum of 4 e at e/4 5 4) 95 items 42) dr 80 = 00 + dx x ) ln 44) e y(y + 6)3 (y - 6) 6 45) y = 7x 5-46) 2 47) 6t5 (t 5 + 3) + 5t 4 (t 6-5) (t 6-5)(t 5 + 3) 48) 4x - 8 (x + 3)( - x) 49) ) - ln 24 (24-x) 6