Review Problems for Test 1

Transcription

1 Review Problems for Test Math 6-03/06 9 9/0 007 These problems are provided to help you study The presence of a problem on this handout does not imply that there will be a similar problem on the test And the absence of a topic does not imply that it won t appear on the test Differentiate fx) = 8x 5 + 4x 9x + Differentiate fx) = 4 x 3 x 3 Differentiate y = 3 7 x Differentiate gu) = u 5 u 4 + u 3 + u + u + ) 4 5 Differentiate hx) = x + x x ) x 6 + x 3 + x + ) 6 Differentiate y = x6 + x 3 + x 4 + x + 7 Differentiate fx) = x ) ) 8 Differentiate fx) = sin 3x + tan5x) 50 9 Differentiate gt) = sincostan t)) + sec 0 Differentiate y = sincos x) + sin x)cos x) Find d x x + 4 ) x + 3 Find d 3x x ) x ) t tan t 3 Compute d [ cos3x + ) + cos3x + )) ] 4 Compute d 6x sin x 7 5 Find d 3 sin x 7x)4 + x tan x) 6 Find f t) if ft) = 3 t + 3t + ) 5 7 Compute d e3x + )sin x)x 4 + x + ) 8 Find d ) x + sec x 3 + 5x cos x 9 Find y if y = 3x x5 )e 4x tan5x + 7

2 0 Compute d + cscx ) e 7x + 3x)7x 6 + x + ) Compute d x ) + x + x 3 3x + 5 Compute d e x + 7 x + e 7x + 7 ex) 3 Compute d 3 x + e x )lnx + 4x) + x 3 4 Compute d dw ln e w + e w + 5 Compute d 6 Compute d x 3 + x 3 + x 3 + ) 3) 3 ) 3 ) x + 5 x + ) 3 7 Compute the derivative with respect to t of st) = sec 8 Compute y for y = x + ) 4 9 Compute y 50) if y = 4 x 30 Compute d e fx) + 3 Compute d lnsingx)) + ) 5 t 3 + 5t f and g are differentiable functions The values of f, g, f, and g at x = and x = 5 are shown below: Compute fg) 5), ) x fx) gx) f x) g x) ) f ), and f g) ) g 33 Find the equation of the tangent line to the graph of y = 34 The position of a cheeseburger at time t is given by st) = t 3 t + 36t + ) x at the point, + 4x + 5) 00 Find the values) of t for which the velocity is 0 Find the values) of t for which the acceleration is 0 35 What is the domain of the function fx) = x + 5x + 6? 36 Let { x 7x 8 fx) = if x x + 9 if x =

3 Prove or disprove: f is continuous at x = 37 Suppose fx) = x + 3 Use the it definition of the derivative to compute f x) 38 Suppose fx) = x + Use the it definition of the derivative to compute f x) 39 The graph of a function fx) is pictured below For what values of x is f continuous but not differentiable? For what values of f is f not continuous? a) Find the average rate of change of fx) = x + 3x 4 on the interval x 3 b) Find the instantaneous rate of change of fx) = x + 3x 4 at x = 3 4 Give an example of a function fx) which is defined for all x and a number c such that fx) fc) x c 4 Compute x 4 x x x + x 3 43 Compute x 3 x 9 44 Compute x 3 x + x 3 x 9 3x 7 5x 4 + x 45 Compute x + 0 x 4 3x 7 3x 7 5x 4 + x 46 Compute x + 0 x 4 3x 6 x 47 Compute x x 5 x Compute x 7 3x + 4 sin 3x 49 Compute x 0 tan 5x x cosx 3

4 50 Suppose { 6 if x 3 fx) = x 9 x 3 if x > 3 Determine whether x 3 fx) is defined If it is, compute it 5 The picture below shows that graph of a function y = fx) Compute: a) fx) b) + fx) c) fx) d) x 3 fx) e) x 3+ fx) f) x 3 fx) g) f3) 5 Locate the horizontal asymptotes and the vertical asymptotes of y = x x 3 5x 6x Justify your answer by computing the relevant its 53 Find the value of c which makes the following function continuous at x = : fx) = { + 3x cx if x < 7x + 3 if x 54 Suppose f is continuous, f) = 3, and f5) = Prove that there is a number x between and 5 such that x 3 fx) = 0 4

5 55 Suppose that gx) = 3, hx) = 9, and x + 8)gx) fx) x + )hx) for all x Compute fx) 56 Find the equation of the tangent line to the curve x 3 + xy 3 = x y + 3 at the point, ) 57 Find y at the point, ) on the curve x 3 y + x = y 3 y Solutions to the Review Problems for Test Differentiate fx) = 8x 5 + 4x 9x + f x) = 40x 4 + 8x 9 Differentiate fx) = 4 x 3 x f x) = x 4 + x 3/ 3 Differentiate y = 3 7 x y = 6 7 x 5/7 4 Differentiate gu) = u 5 u 4 + u 3 + u + u + ) 4 g u) = 4u 5 u 4 + u 3 + u + u + ) 4 5u 4 4u 3 + 3u + u + ) 5 Differentiate hx) = x + x x ) x 6 + x 3 + x + ) h x) = x + x x ) ) 6x 5 + 3x + ) + x 6 + x 3 + x + ) x / + x 6 Differentiate y = x6 + x 3 + x 4 + x + y = x4 + x + )6x 5 + 3x ) x 6 + x 3 + )4x 3 + x) x 4 + x + ) 5

6 7 Differentiate fx) = x ) ) f x) = ) x ) ) 3 ) 3 + x ) 3 )x 3 8 Differentiate fx) = sin 3x + tan5x) 50 f x) = 50sin 3x + tan 5x) 49 3 cos3x + 5sec 5x) ) ) t 9 Differentiate gt) = sincostan t)) + sec tan t ) t t g t) = coscostant)) sintant)) sec t) + sec tan tant tan t ) tant)) t)sect) tant) ) 0 Differentiate y = sincos x) + sin x)cos x) y = coscos x)) sin x) + sin x) sinx) + cosx)cosx)) Find d x x + 4 ) x + 3 Note that Then x x + 4 x + 3 = x x + 4x + 3 ) d x x + 4x + 3 = 7 3 x x 3 = 7 3 x x 3 Find d 3x x ) x Note that 3x x + 7 x = 3 x + 7x /3 + x /7 = 3 x / + 7x /3 + x /7 Then d 3 x / + 7x /3 + x /7) 3 = x / 7 3 x 4/3 + 7 x 5/7 3 Compute d [ cos3x + ) + cos3x + )) ] d [ cos3x + ) + cos3x + )) ] = sin3x + ) ))3x + )3) + cos3x + )) sin3x + ))3) = 6

7 63x + ) sin3x + ) 6 cos3x + ) sin3x + ) 4 Compute d 6x sin x 7 d 6x sin x = d 7 7 6x sin x ) = 8x + 3 cosx ) 7 It is not a good idea to use the Quotient Rule when either the top or the bottom of a fraction is a number 5 Find d 3 sin x 7x)4 + x tan x) d 3 sin x 7x)4 + x tan x) = 3 sinx 7x)x secx) ) x tanx)3 cosx 7) 6 Find f t) if ft) = Rewrite the function as Then 3 t + 3t + ) 5 ft) = 3t + 3t + ) 5 f t) = 3) 5)t + 3t + ) 6 5t + 3) t + 3) = t + 3t + ) 6 It is not a good idea to use the Quotient Rule when either the top or the bottom of a fraction is a number 7 Compute d e3x + )sin x)x 4 + x + ) For a product of three terms, the Product Rule says So d st ) nd )3 rd ) = st ) nd ) []derx3 rd ) ) ) ) d d + st ) nd ) 3 rd ) + st ) nd )3 rd ) d e3x + )sinx)x 4 + x + ) = e 3x + )sin x)4x 3 + ) + e 3x + ) cosx)x 4 + x + )+ 3e 3x )sinx)x 4 + x + ) 8 Find d ) x + sec x 3 + 5x cos x Note that x + secx 3 + 5x cos x = x/ + sec x 3 + 5x cos x 7

8 Then ) d x / ) 3 + 5x cos x) + sec x x / + sec x tanx ) x / + sec x)5 + sinx) = 3 + 5x cos x 3 + 5x cos x) = 3 + 5x cos x) x ) + sec x tanx x + secx)5 + sin x) 3 + 5x cos x) 9 Find y if y = 3x x5 )e 4x tan5x + 7 y = tan5x + 7) 3x x 5 )4e 4x ) + e 4x ) 3 5x 4 ) ) 3x x 5 )e 4x 5)sec5x) tan5x + 7) I applied the Quotient Rule to the original fraction In taking the derivative of the top, I also needed to apply the Product Rule 0 Compute d + cscx ) e 7x + 3x)7x 6 + x + ) d + cscx ) e 7x + 3x)7x 6 + x + ) = e 7x + 3x)7x 6 + x + )x) cscx ) cotx )) + cscx )) e 7x + 3x)4x 5 + ) + 7x 6 + x + ) 7e 7x + 3) ) e 7x + 3x) 7x 6 + x + ) I applied the Quotient Rule to the original fraction In taking the derivative of the bottom, I also needed to apply the Product Rule Compute d x ) + x + x 3 3x + 5 d x ) + x + x + x + x 3 = 3x + 5 x 3 3x + 5 ) 0 x 3 3x + 5)x + ) x + x + )3x 3) x 3 3x + 5) ) Compute d e x + 7 x + e 7x + 7 ex) d e x + 7 x + e 7x + 7 ex) = xe x + xln7)7 x + 7e 7x + e x )ln7)7 ex 3 Compute d 3 x + e x )lnx + 4x) + x 3 Note that 3 x + e x )lnx + 4x) + x 3 = x + e x )lnx + 4x) + x 3) /3 8

9 Then x + e x )lnx + 4x) + x 3) /3 3 d x + e x )lnx + 4x) + x 3) /3 = ) ) x + e x ) x lnx + 4x)x + e x ) + 3x 4 Compute d dw ln e w + e w + Note that Then ) / ln e w + e w + = ln e w + e w ) + = ln e w + e w + d ) ) dw ln e w + e w + = wew + e w e w + e w + = wew + e w e w + e w + 5 Compute d x 3 + x 3 + x 3 + ) 3) ) 3 3 d x 3 + x 3 + x 3 + ) 3) ) 3 3 = 3 x 3 + x 3 + x 3 + ) 3) ) 3 3x + 3 x 3 + x 3 + ) 3) 3x + 3x 3 + ) 3x ) )) 6 Compute d ) x + 5 x + ) 3 ) d x + 5 x + ) 3 = x + )3 ) x + 5)3)x + ) x + )) x + 5)3) x + ) 6 = x + ) 4 = x + 4 6x 5 4x x + ) 4 = x + ) 4 In going from the second expression to the third, I cancelled a common factor of x + ) If you try to multiply before you cancel, you ll get a big mess, and it will be much harder to simplify 7 Compute the derivative with respect to t of st) = sec Rewrite the function as Then 5 t 3 + 5t + 8 st) = sec 5t 3 + 5t + 8) ) s t) = sec 5t 3 + 5t + 8) ) tan 5t 3 + 5t + 8) )) 5) )t 3 + 5t + 8) 3t + 5) ) = sec 5t 3 + 5t + 8) ) tan 5t 3 + 5t + 8) )) 53t ) + 5) t 3 + 5t + 8) ) 9

10 8 Compute y for y = x + ) 4 y = 4x + ) 3 x) = 8xx + ) 3, y = 8 x)3)x + ) x) + x + ) 3) = 48x x + ) + 8x + ) 3, y = 48 x ))x + )x) + x + ) x) ) + 8)3)x + ) x) = 9x 3 x + ) + 96xx + ) + 48xx + ) = 9x 3 x + ) + 44xx + ) 9 Compute y 50) if y = 4 x I ll compute the first few derivatives until I see the pattern y = y = 4 x), 4 x) 3, y = 3 4 x) 4, y 4) = x) 5 Note that I don t get minus signs here; the powers are negative, but the Chain Rule requires the derivative of 4 x, which is The two negatives cancel Note also that, in order to see the pattern, I did not multiply out the numbers on the top Based on the pattern, I see that the power on the bottom is one more than the order of the derivative On the top, I get the product of the numbers from through the order of the derivative So You can also write the top as 50! 50-factorial)) y 50) = x) 5 30 Compute d e fx) + Since fx) is not given, the answer will come out in terms of fx) and f x) Note that / e fx) + = e fx) + ) Then d ) / ) / e fx) + = e fx) + e fx)) fx))f x)) 3 Compute d lnsingx)) + ) Since gx) is not given, the answer will come out in terms of gx) and g x) d lnsingx)) + ) = cosgx))g x)) singx)) + 0

11 3 f and g are differentiable functions The values of f, g, f, and g at x = and x = 5 are shown below: x fx) gx) f x) g x) Compute fg) 5), By the Product Rule, By the Quotient Rule, By the Chain Rule, ) f ), and f g) ) g fg) 5) = f5)g 5) + g5)f 5) = 05) 3) + )7) = 85 ) f ) = g)f ) f)g ) 5)4) 6)0) g g) = 5 = 6 5 f g) ) = f g)) g ) = f 5) g ) = 7 0 = Find the equation of the tangent line to the graph of y = y = x + 4x + 5), so y = )x + 4x + 5) 3 x + 4), y ) = 3 50 ) x at the point, + 4x + 5) 00 The tangent line is 3 x ) = y The position of a cheeseburger at time t is given by st) = t 3 t + 36t + Find the values) of t for which the velocity is 0 Find the values) of t for which the acceleration is 0 The velocity is the derivative of the position: vt) = s t) = 3t 4t + 36 = 3t 8t + ) = 3t )t 6) I have vt) = 0 for t = and t = 6 The acceleration is derivative of the velocity or the second derivative of the position): at) = v t) = 6t 4 = 6t 4)

12 Note: I differentiated vt) = 3t 4t+36, not vt) = 3t )t 6) Differentiating the first expression is easier than differentiating the second) I have at) = 0 for t = 4 35 What is the domain of the function fx) = fx) = x + )x + 3) x + 5x + 6? I can t divide by 0, and division by 0 occurs where x + )x + 3) = 0 Square both sides: x+)x+ 3) = 0 I get x = or x = 3 I can t take the square root of a negative number, and this occurs where x + )x + 3) < 0 Solve the inequality by graphing the quadratic y = x + )x + 3) is a parabola opening upward since it s y = x + 5x + 6) It has roots at x = and at x = 3 Picture: -3 - x + )x + 3) < 0 for 3 < x < I throw out the bad points x =, x = 3, 3 < x < The domain is x < 3 or x > 36 Let { x 7x 8 fx) = if x x + 9 if x = Prove or disprove: f is continuous at x = x 7x 8 x 8)x + ) fx) = = = x 8) = 9 x x x + x x + x On the other hand, f ) = 9 Since fx) f ), f is not continuous at x = x 37 Suppose fx) = x + 3 Use the it definition of the derivative to compute f x) Therefore, fx + h) fx) = fx + h) fx) h = x + h + 3 x + 3) x + h + 3) h = = x + 3 x + h + 3)x + 3) x + h + 3)x + 3) h x + h + 3)x + 3) h = h hx + h + 3)x + 3) = x + h + 3)x + 3)

13 Hence, f fx + h) fx) x) = = h 0 h h 0 x + h + 3)x + 3) = x + 3) 38 Suppose fx) = h 0 h 0 x + Use the it definition of the derivative to compute f x) f fx + h) fx) x) = = h 0 h h 0 x + h + x + h 0 h x x + h h x + ) x + h + ) = h 0 x x + h) h x + ) x + h + ) x + x + h) = h 0 h 0 x + h + h x + x + ) x + h + ) x + ) x + h + ) = x + ) x + h + ) h 0 h x + ) x + h + ) = x x + h x + x + h h x + ) x + h + ) = x + x + h h h x + ) x + h + ) x + x + h) = x + ) x + h + ) x + x + h) = x + ) x = 39 The graph of a function fx) is pictured below For what values of x is f continuous but not differentiable? For what values of f is f not continuous? f is continuous but not differentiable at x = and at x = 7, since at those points the graph is unbroken but has corners f is not continuous at x = 4 and at x = 40 a) Find the average rate of change of fx) = x + 3x 4 on the interval x 3 f3) f) 3 = = 7 3

14 b) Find the instantaneous rate of change of fx) = x + 3x 4 at x = 3 f x) = x + 3, so f 3) = = 9 4 Give an example of a function fx) which is defined for all x and a number c such that fx) fc) x c There are lots of possible answers to this question For example, let { fx) = x if x 0 if x = 0 Then fx) = x = 0, but f0) = x 0 x 0 Of course, the condition x c fx) fc) means that the function is not continuous at c The value of x c fx) or even its existence) does not depend on the value of fc) or even its existence) 4 Compute x 4 x x x 4 x x = x )x + ) x )x + ) = x + x + = 4 3 Since plugging in x = gives 0, it is reasonable to suppose that the zeros are being caused by a common 0 factor on the top and the bottom So factor the top and the bottom and cancel the x s After that, plugging in x = a second time gives 4 3 x + x 3 43 Compute x 3 x 9 x + x 3 x 3 x is undefined 9 If I plug in x = 3, I get, a nonzero number over 0 In this case, the it does not exist 0 By the way, I hope you didn t try to use L Hôpital s Rule here It doesn t apply) 44 Compute x 3 x + x 3 x 9 x + x 3 x 3 x = 9 This problem is different from the previous one because x is approaching 3 from the left It would not be incorrect to say that the it is undefined, but it is better to say that the it is There is a vertical asymptote at x = 3, and the graph goes downward to ) as it approaches the asymptote from the left How do you see algebraically that this is what it does? One way is to plug numbers close to 3, but less than 3, into x + x 3 x For example, if x = 99999, x + x 3 9 x a big negative number 9 This doesn t prove that it s going to, but it s pretty good evidence 4

15 Here is how I can analyze it I know that plugging in x = 3 gives Therefore, I should be getting 0 either + or the reciprocal of something small 0) should be something big Now let x approach 3 from the left x + x 3 surely goes to, a positive number x 9 goes to 0, but since x < 3, x 9 will be negative Now positive = negative, so the answer is negative 3x 7 5x 4 + x 45 Compute x + 0 x 4 3x 7 3x 7 5x 4 + x 3 5 x + 0 x 4 3x 7 = x 3 + x 6 x x + 0 x 7 = = x Divide the top and bottom by the highest power of x that occurs in either in this case, x 7 Then use the fact that number = 0 x ± xpositivepower 3x 7 5x 4 + x 46 Compute x + 0 x 4 3x 6 3x 7 5x 4 + x 3 5 x + 0 x 4 3x 6 = x 3 + x 6 x 7 x + 0 x 7 x 3 3 x = = There s little question that the it is undefined; why is it? One approach is to note that the top and bottom are dominated by the highest powers of x So this its looks like x + 3x 7 3x 6 = x = x + As a check, if you plug x = 0 6 into 3x7 5x 4 + x 0 x 4 3x 6, you get approximately) 0 6 x 47 Compute x x x x x ) x + ) = = x + ) = + = x x x x Plugging in x = gives 0 As usual, I look for a common factor that is producing the zeros Since the 0 bottom is smaller than the top, it s reasonable to see if the bottom is actually a factor or the top It is; I cancel the x s, and plug in to finish 5 x Compute x 7 5 x + 5 x 7 =

16 Nothing going on here I just plug in x = By the way Do you know how to compute on your calculator?) 3x + 4 sin 3x 49 Compute x 0 tan 5x x cosx x 0 3x + 4 sin 3x tan 5x x cosx = x 0 3x + 4 sin 3x = sin 5x cos5x x cosx x 0 3x + 4 sin 3x sin 5x cos5x x cosx x x = x 0 sin 5x x sin 3x x = cos5x cosx sin 3x 3x = x 0 sin 5x 5 5x cos5x cosx = Suppose { 6 if x 3 fx) = x 9 x 3 if x > 3 Determine whether x 3 fx) is defined If it is, compute it Since { 6 if x 3 fx) = x 9 x 3 if x > 3 is made of two pieces glued together at x = 3, I ll compute x 3 fx) by computing the its from the left and right, and seeing if they re equal If they aren t equal, the it is undefined) The left-hand it is The right-hand it is fx) = 6 = 6 x 3 x 3 x 9 fx) = x 3+ x 3+ x 3 = x 3)x + 3) = x + 3) = 6 x 3+ x 3 x 3+ Since the left- and right-hand its are equal, x 3 fx) is defined, and fx) = 6 x 3 6

17 5 The picture below shows that graph of a function y = fx) Compute: a) fx) b) + fx) c) fx) d) x 3 fx) e) x 3+ fx) f) x 3 fx) g) f3) fx) = 5 fx) = 4 + fx) is undefined + fx) = x 3 fx) = x 3+ fx) = x 3 f3) = 3 5 Locate the horizontal asymptotes and the vertical asymptotes of y = x x 3 5x 6x Justify your answer by computing the relevant its Since x x x 3 5x = 0 and 6x x y = 0 is a horizontal asymptote at + and at 7 x x 3 5x 6x = 0,

18 Factor the bottom: y = x xx 6)x + ) The function is undefined at x = 0, x = 6, and at x = Note that x x 0 xx 6)x + ) = x x 0 x 6)x + ) = 0 Hence, the graph does not have a vertical asymptote at x = 0 Consider the its at x = 6 I know the one-sided its will be either + or, since plugging in x = 6 gives 36 0, a nonzero number divided by 0 Take x 6 + as an example x 36, x 6, x + 7, and x 6 0 In the last case, since x > 6, x x 6 > 0, ie x 6 goes to 0 through positive numbers Since all the factors of are positive xx 6)x + ) as x 6 +, the quotient must be positive, and the it is + Similar reasoning works for the left-hand it Thus, x x = + and x 6 + xx 6)x + ) x 6 + xx 6)x + ) = There is a vertical asymptote at x = 6 Likewise, x x = + and x + xx 6)x + ) x + xx 6)x + ) = There is a vertical asymptote at x = The results are visible in the graph of the function: Find the value of c which makes the following function continuous at x = : fx) = { + 3x cx if x < 7x + 3 if x For f to be continuous at x =, fx) must be defined This will happen if the left and right-hand its at x = are equal Compute the its: fx) = + +7x + 3) = 7, fx) = + 3x cx ) = 7 4c 8

19 Set the left and right-hand its equal and solve for c: 7 4c = 7, 4c = 0, c = 5 This value of c will make the left and right-hand its equal to 7, so in this case, fx) = 7 Since f) = 7, it follows that fx) = f), and f is continuous at x = 54 Suppose f is continuous, f) = 3, and f5) = Prove that there is a number x between and 5 such that x 3 fx) = 0 x 3 and fx) are continuous, so x 3 fx) is continuous When x =, x 3 fx) = 8 3 = 04 When x = 5, x 3 fx) = 5 = 5 Since x 3 fx) is continuous, and since 0 is between 04 and 5, the Intermediate Value Theorem says that there is a number x between and 5 such that x 3 fx) = 0 55 Suppose that gx) = 3, hx) = 9, and x + 8)gx) fx) x + )hx) for all x Compute fx) x + 8)gx) = 3 = 36 and x + )hx) = 4 9 = 36 By the Squeezing Theorem, fx) = Find the equation of the tangent line to the curve x 3 + xy 3 = x y + 3 at the point, ) Differentiate implicitly: Plug in x =, y = : The tangent line is 3x + y 3 + 6xy y = x yy y = y, y = x ) = y 8 57 Find y at the point, ) on the curve Differentiate implicitly: x 3 y + x = y 3 y x 3 y + 3x y + 4x = 3y y y ) Plug in x = and y = and solve for y : y = y y, y = 0 9 9

20 Differentiate *) implicitly: x 3 y + 3x y + 3x y + 6xy + 4 = 3y y + 6yy ) y Plug in x =, y =, and y = 0 9 and solve for y : y + 3 ) ) = y + 9 ) 0 y, y = 9 43 Only the brave know how to forgive A coward never forgave; it is not in his nature - Laurence Sterne c 008 by Bruce Ikenaga 0