There are 9 problems, most with multiple parts. The Derivative #1. Define f: R\{0} R by [f(x) = 1/x] Use the definition of derivative (page 1 of Differentiation notes, or Def. 4.1.1, Lebl) to find, the derivative of f at c. f ( c ) = lim{x->c} (f(x)-f(c))/(x-c) = lim{x->c} (1/x - 1/c)/(x-c) = lim {x->c} ( (c - x)/( c x)) / (x-c) = lim {x->c} -1/( c x) = - 1 / x #. Let Use the definition of derivative (page 1 of Differentiation notes, or Def. 4.1.1, Lebl) and the result of Week 5 Homework #1(b) to find, the derivative of f at 0. lim{x->0} (f(x) - f(0))/(x - 0) = lim{x->0} (x sin(1/x) - 0)/x (as the limit doesn t allow x to be 0) = lim{x->0} x sin(1/x) = 0 as x sin(1/x) <= x, so delta = epsilon works in the definition of limit. Page 1 of 11
#3. Is differentiable at 1? Explain carefully and justify your answer. No. If x>1, (f(x) - f(1))/(x-1) = (3 x - (1 + 1)) / (x-1) = (3 x - )/(x-1) Where x = 1 + (1/n), (3 x - )/(x - 1) = (1 + (1/n))/(1/n) = n+1, so lim{x->1} (f(x) - f(1))/(x-1) does not exist. #4. Fill in the blanks to use the Mean Value Theorem and the known facts that and to carefully show that, i.e., Proof: [Indicate what belongs in each of the blanks) Let on the interval [36, 39]. Since f is a continuous function on [36, 39], differentiable on (36, 39) the Mean Value Theorem can be applied, and so there exists c _(36,39) (state the interval) such that Now we want a lower bound and an upper bound for. f ( c) = = _1/( sqrt(c)). (State the derivative, in terms of c.) Page of 11
Since c > 36, f ( c) < _ 1/( sqrt(36)) = 1/1. (State an appropriate fraction, a "nice" rational number.) Since c < 39 < 49, f ( c) > 1/( sqrt(49)) = 1/14. (State an appropriate fraction, a "nice" rational number. Make use of the fact that we know ; we don't know an approximation for ; indeed that is what we are trying to determine). Then 1/14 < f ( c) < 1/1 (leave as fractions, in lowest terms). Thus 6 + 3/14 < 6 + 3 f ( c) < 6 + 3/1 = 6 + 1/4 which is what we are trying to show. Page 3 of 11
You may assume relevant derivative formulas from calculus for powers of x, polynomials, sine, cosine, tangent, exponentials, and logs. #5. Let f(x) = 4x + 4x + 9 for x in [0, ]. #5 (a) Use the derivative to find where f is increasing and where f is decreasing. (Show some work.) f (x) = - 4 x + 4 = -8x + 4 = -8(x - ½), so f (x) is positive for x < ½, and negative for x > ½. Hence f is increasing on [0, ½], and decreasing on [½, ] #5 (b) Find the maximum and the minimum of f on [0, ]. Show work/explanation. Because f is increasing on [0, ½], and decreasing on [½, ], the maximum of f must occur at x = ½, with f(½) = 10 As a local minimum of f must occur at a point c where f ( c) = 0, the minimum of f must occur at an endpoint of the interval. Calculating f(0) = 9, f() = 1, so the minimum occurs at x =, and f() = 1. #6. For each of the following scenarios, find a function f and a domain [a, b] which satisfies the scenario and for which there is NO point c in (a, b) with f (c) = 0. (You can have a different function and interval for each part). By carrying out this exercise, you are verifying that all of the hypotheses for Rolle's Theorem must be satisfied in order to guarantee the conclusion that f (c) = 0 for some c in (a, b). (Explanations not required) #6(a) f is continuous on [a, b] and f is not differentiable on (a, b) and f(a) = f(b), and there is no point c in (a, b) with f (c) = 0. f(x) = x on [-1,1]; #6 (b) f is continuous on [a, b] and differentiable on (a, b) and f(a) f(b), and there is no point c in (a, b) with f (c) = 0. f(x) = x on [0,1] # 6(b) f is not continuous on [a, b] and f is differentiable on (a, b) and f(a) = f(b), and there is no point c in (a, b) with f (c) = 0. f(x) = [begin cases] 0 if x=-1 or 1 x if -1 < x < 1 Page 4 of 11
on [-1,1] Page 5 of 11
L'Hopital's Rule You may assume relevant derivative formulas from calculus for powers of x, polynomials, sine, cosine, tangent, exponentials, and logs. #7. Determine the following limits (if they exist). Show work. [Note: some symbols do not display on any of the programs available to me. They might change some of the answers] #7(a). Both the numerator (tan 0) and the denominator (sin 0) are 0 at the limit point, so L Hopital s Rule can be applied lim{x->0} tan 9x / sin 4x = lim{x->0} (d/dx) tan 9x / (d/dx) sin 4x L Hopital s rule, limits 0 = lim {x->0} 9 csc 9x / 4 cos 4x =( lim {x->0} 9 csc 9x ) / (lim {x->0} 4 cos 4x) as the limits exist = (9 csc 0) / (4 cos 0) As csc and cos are continuous at 0 = 9/4 #7(b). lim {x-> } (x ln x)/(exp(3x)) = lim{x-> } ((d/dx) (x ln x))/((d/dx) exp(3x)) L Hopital s rule, limits = lim{x-> } (1 + ln x)/(3 exp(3x)) = lim{x-> } ((d/dx) (1 + ln x)) /((d/dx) (3 exp(3 x))) L Hopital s rule, limits =lim{x-> } (1/x) /(9 exp(3 x)) = [lim{x-> ) (1/x)] [ lim{x-> ) (1/9)] ]lim{x-> ) exp(-3 x)] as the limits exist = 0 (1/9) 0 = 0 Page 6 of 11
#7(c). Hints:. First find Per hint; lim {x-> } x ln (1-4/x) = lim {x-> } ln(1-4/x)/(1/x) reorder = lim {x-> } [(d/dx) ln (1-4/x)] /[(d/dx) (1/x)] L Hospital s rule, limits 0 = lim {x-> }[ (4/x )/(1-4/x)]/(-1/x ) = lim {x-> } -4 / (1-4/x) Simplify =( lim {x-> } -4 )/(lim {x-> } (1-4/x)) limits exist and are nonzero = -4 / 1 = -4 Alternatively, lim {x-> } x ln (1-4/x) = lim{y->0} (1/y) ln (1-4 y) Substitute y = 1/x; x = 1/y = lim{y->0} (ln (1-4y)/y) Reorder = lim{y->0} ( (d/dy) ln (1-4y))/ ((d/dy) y) L Hospital s rule, limits 0 = lim{y->0} (-4/(1-4 y))/1 = (lim {y->0} -4)/(lim{y->0} 1-4y } limits exist and are nonzero = -4/1 = -4 x Now, lim {x-> } (1-4/x) = lim {x-> } exp( x ln (1-4/x)) Per hint = exp (lim {x-> } x ln(1-4/x)) exp is continuous -4 = exp(-4) = e limit calculated above Page 7 of 11
#8. Critique the following work. Is it correct? If not, explain what has been done wrong, and correctly determine the limit. The first equality is a correct application of L Hospital s rule, as 3 3 lim{x->} x - x + x - = - ( ) + - = 0, and 3 3 lim{x->} x -x = -( ) = 0 However, the second application is not as lim{x->} 3x -4x + 1 = 3( )-4() + 1 = 5 lim{x->} 3x -4x = 3( )-4() = 4 So the limit is 5/4. Infinite Series #9. For each of the following six series, decide if it is convergent or divergent. Justify your answer with explanation/work. Reference appropriate examples, theorems, or tests from the Week 6 Infinite Series notes or Section.5 of Lebl. #9(a) Divergent by the ratio test; a_{n+1} / a_{n} = e/ > 1 #9(b) Divergent by the limit test (Proposition.5.8) lim {n > } (sqrt(n +1)/(3n -1)) = lim{n-> } ( (n sqrt(1 + 1/n ))/(n (3-1/n)) =lim{n-> } ( sqrt(1 + 1/n )/(3-1/n)) Page 8 of 11
= 1/3 notequal 0 Page 9 of 11
#9 (c) Convergent by the ratio test n+1 n a_{n+1} / a_n = ((n+1) /3 )/ (n /3 ) = ((n+1)/n) /3= (1+1/n) /3 lim {n > } a_{n+1} / a_n = lim {n > } (1+1/n) /3 = 1/3 < 1 #9(d) convergent by the ratio test (actually, it s a geometric series) n 4n n n n a_n = (8 )/(3 ) = (64 )/(81 ) = (64/81) a_{n+1} / a_n = 64/81 < 1 #9 (e) Convergent by the ratio test n+1 n a_{n+1}/a_{n} = (100 )/((n+1)!) / ((100 )/n!) = 100/(n+1) lim {n > } a_{n+1}/a_{n} = 0 Page 10 of 11
#9 (f) Convergent by the ratio test a_{n+1}/a_n = ((n+1) (0.9) )/(n (0.9) ) = ((n+1)/n) (0.9) = (1 + 1/n)(0.9) n+1 n Page 11 of 11