y+2 x 1 is in the range. We solve x as x =

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1 Dear Students, Here are sample solutions. The most fascinating thing about mathematics is that you can solve the same problem in many different ways. The correct answer will always be the same. Be creative and have fun.. The solutions of x satisfy the following inequality x is (a) [2, ), (b). (, 2], (c) [ 2, 2], (d) [ 6, 6], (e) None of the above. Remember that more than one answer satisfies c 2 = b ie:( 2) 2 = 4 and 2 2 = 4. So adding 4 to both sides gives x 2 4 so this is true when x 4 and when x 4 ie: when x [ 2, 2]. 2. Given f(x) = 2x 3 5x + 4, then f( ) is: (a) (b), (c) 7, (d)3, (e) None of the above. f( ) = 2( ) 2 5( ) + 4 = = 7 3. The domain and range of the function g(x) = x + 2 x is: (a) [ 2, ), (b). (, 2], (c) {x : x }, (d) {x : x 2}, (e) (, 2]. The domain of g(x) is every x so that g(x) makes sense. Since g(x) does not make sense when the denominator is zero we see that the domain (and also the range) are {x x } To find the range, let y = x+2 y+2 x is in the range. We solve x as x = y. THat is for all y, this is an x that hits y. Hence the range is {x }. 4. The equation of the line joining the points (, -) and (3, 3). (a) y = 3x + 2.(b) y = 2x 3, (c) y = x, (d)x + y = 4, (e) None of the above. First find the slope. m = 3 ( ) Then use the point-slope equation and the fact that 3 (3,3) is a point on the line to get y 3 = 2(x 3) where the slope, m, is m = 2 (calculated above). Simplifying gives the equation of the line is y = 2x 3 5. Which is the domain of the function f(x) = x 3? a) [ 3, ), (b). (, 3], (c) {x : x 3}, (d) {x : x 3}, (e) (, 0]. We see that f(x) does not make sense if what is under the radical is negative. So we need x 3 0 so that the domain is {x x 3} = [3, ) 6. Given f(x) = x 2 and g(x) = x, find: (f g)(5) is (a), (b) 2, (c) 3, (d)4, (e) None of the above. f(g(x)) = ( x ) 2 = x so f(g(5)) = 5 = Given f(x) = x 2 and g(x) = x, find: (g f)(5) is (a), (b) 2, (c) 3, (d)4, (e) None of the above.

2 g(f(x)) = x 2 so g(f(5)) = 25 = The average rate of change of g(x) =, x = 2, x = 0. is x (a) etc. Average rate of change is g(x + x) g(x) x so with x=2 and x =. we have = 9. The average rate of change of f(x) = x 2, at with x = 0. is (a), etc. f(x + x) f(x) Average rate of change is so with x= and x =., we have ( +.)2 2 x The it of x 2 + x 2 x + 2 = Answer: 5/4. The it of x 2 x 2 x 2 x 2 Answer: 3 2. The it of x 2 + x 2 x 2 x + 2 Answer: The it of x 2 x 2 x x 2 Answer: 2 4. Which is the it + x x 0 x (a) 0, (b), (c) 2, (d) 0.5 (e), -. Answer: d 2

3 5. Which is the it 4 x 2 x 0 x (a) 0, (b) -, (c), (d) -0.5 (e), Answer: e 6. What is the following one-sided it: x x x (a) 0, (b) -, (c), (d) -0.5 (e), doesn t exist. Answer: b 7. Consider the function h(x) = 3x+ x+ and g(x) = x 2. Which of the following is true. (a) f and g are continuous at 2, (b) Only f is continuous at 2,(c) Only g is continuous at 2,(d) Neither is continuous at 2,(e)None of the above. Answer: a 8. What is the vertex of the parabola y = 2x 2 x + 3. Answer: (/4,23/8) 9. What is the vertex of the parabola y = 2x 2 x + 3. Solve by completing the square Therefore the vertex is at ( 4, 25 8 ). y = 2x 2 x + 3 = 2(x x) + 3 = 2(x x ) + 3 = 2(x + 4 ) The tangent line of 2+x at (2, 2 ). Determine the slope by computing the derivative at x = 2. and so Therefore the equation of the tangent line is; f (x) = 2 (2 + x) 3 f (2) = 6. y 2 = (x 2) 6 3

4 or y = 6 x The derivative of the function y = 3 2x 2 is (a) (b) (c) (d). y = 4x 22. What is 2. 9? For this and next two questions, the best is to use the calculator get the answer right away. You could use some of the following tricks, but won t help much = ( )9 = = What is ? 24. What is ln( )? = ( 5 2 )5 = = 32 0 ln( ) = 5 2 ln 5 2 = 5 (ln 5 ln 2) What is log 2 (3.5)? ln 3.5 log 2 (3.5) = ln What is the tangent line of the function + x at the point (0, )? So the tangent line at (0, ) is d + x = ( + x) /2 dx 2 y = (x 0) 2 or y = 2 x What is the tangent line of the function ln( + x) at the point (0, 0)(the original has a typo, the point should be on the graph of the function)? d ln( + x) = dx + x So the tangent line at (0, 0) is or y = (x 0) y = x 28. What is the tangent line of the function 2 +2x at the point (0, 2)? f(x) = 2 +2x = e ln(2+2x) = e (+2x)(ln(2)) So use the chain rule to get that f (x) = 2ln2 e (+2x)(ln(2)) f (0) = 2ln2 e ( ln(2)) = 4ln2 Using point-slope form for the line we obtain y 2 = 4ln2 x 4

5 29. The derivative of (x 2 ) 5 is. Use the chain rule, f(x) = x 5, g(x) = x 2 to obtain (f g(x)) = f (g(x))g (x) = 5(x 2 ) 4 (2x) = 0x(x 2 ) The derivative of x x 2 + is. Use the product rule and the chain rule to obtain (x x 2 + ) = (x) x 2 + +(x)(sqrtx 2 + ) = () x 2 + +x( 2 (x2 +) x 2 (x 2 +) 3. The derivative of x2 x+ is. Use the quotient rule to obtain ( x2 x+ ) = (x2 ) (x+) (x 2 )(x+) = 2x(x+) (x2 +)() = 2x2 +2x x 2 + (x+) 2 (x+) 2 (x+) The derivative of (x 2 ) 5 is. This is a repeat problem, see above. 2 2x) = x = (x+)2 (x+) 2 =, x 33. The derivative of e x2 is. Use the chain rule, f(x) = e x, g(x) = x 2 to obtain that (e x2 ) = (f(g(x))) = f (g(x))g (x) = 2xe x2 34. The derivative of e x+2 is. Use the chain rule again, f(x) = e x, g(x) = x + 2 to get (e x+2 ) = (f(g(x))) = f (g(x))g (x) = ( )e x The derivative of xe x+2 is. Use the product rule to get (xe x+2 ) = (x) (e x+2 ) + (x)(e x+2 ) = ()(e x+2 ) + (x)(( )e x+2 ) = e x+2 ( x) 36. The derivative of 2 x3 is. Since 2 x3 = e ln(2 x3 ) = e x3ln(2), use the chain rule with f(x) = e x, g(x) = x 3 ln2 to get that (f(g(x))) = f (g(x))g (x) = (e x3 ln2 )( 3x 2 ln2) = e ln(2 x3 ) ( 3 ln2 x 2 ) = (2 x3 )( 3 ln2 x 2 ) 37. The derivative of f(x) = ln( + x 2 ) is:. 38. The derivative of f(x) = ln( + 2x 2 ) is: f (x) = 2x (2x) = + x2 + x 2. f (x) = 4x (4x) = + 2x2 + 2x 2 5

6 39. The derivative of f(x) = x ln( + 4x) is:. f 4x (x) = x (4) + ln( + 4x) = + ln( + 4x) + 4x + 4x 40. The tangent line of f(x) = x ln( + 4x) at (, ln 5) is: First notice: f () = ln( + 4) = 4 + ln 5. 5 So the equation for the tangent line is: y ln 5 = ( 4 + ln 5)(x ) The tangent line of f(x) = e +4x at (0, e) is found as follows. First, f (x) = e +4x (4) = 4e +4x. So f (0) = 4e = 4e. Then the equation for the tangent line at (0,e) is: y e = 4e(x 0). Simplifying this we get: y = 4ex + e. 42. A sample of a microorganism grows from gram to 5 grams in 24 hours. What is its growth rate in an hour? Solution: First consider the equation y = ae kx. 5 = e 24k ln 5 = ln e 24k ln 5 = 24k ln e ln 5 = 24k k = ln Now our equation is y = ae.067t. So plugging in t= and a= we get: Hence the growth rate in one hour is 7%. y = e A sample of a microorganism grows from 5 grams to 5 grams in 24 hours. What is its growth doubling time? Solution: First consider the equation y = ae kx. 5 = 5e 24k 3 = e 24k ln 3 = ln e 24k ln 3 = 24k k = ln Now our equation is y = ae.0458t. So plugging in y = 2a we get: 2a = ae.0458t 6

7 2 = e.0458t ln 2 = ln e.0458t ln 2 =.0458t t = ln Hence the doubling time is approximately 5.3 hours. 44. A sample microorganism grows from gram to 5 grams in 24 hours. What is its doubling time? Solution: First note that the setup is the same as in # 42 So our equation is y = ae.067t. So plugging in y = 2a we get: 2a = ae.067t 2 = e.067t ln 2 = ln e.067t ln 2 =.067t t = ln Hence the doubling time is approximately 0.3 hours. 45. Suppose C 4 s half life is 5570 years. How many years does it decay to 75%? 2 = e5570k ln 2 = ln e5570k [Note that ln 2 = ln ln 2 = 0 ln 2 = ln 2] ln 2 = 5570k k = ln So to solve the original problem we use.75 = e ln t. ln.75 = ln e ln t ln.75 = ln t t = (ln.75) 5570 ln Hence the time it takes to decay to 75% is about years. 7