Math 19: Calculus Winter 2008 Instructor: Jennifer KlokePractice Final1. Determine whether each statement is true or false. Unless otherwise stated, any func-tion below is arbitrary. If the statement is true, very briefly cite your reasoning.If it is false, provide an example showing the statement to be false.(a) True or False. If f0(x) < 0 for 1 < x < 6, then f(x) is decreasing on (1,6).Solution: True. The f0(x) < 0 if and only if f(x) is decreasing.(b) True or False. If f(x) has an absolute minimum value at x = c, the f0(c) = 0.Solution: False. f(x) = |x| has absolute minimum at x = 0 but f0(0) does notexist.(c) True or False. f0(x) has the same domain as f (x).Solution: False. f(x) = |x| is defined everywhere but f0(x) is not defined whenx = 0.(d) True or False. If f(x) and g(x) are differentiable, thenddx[f(g(x))] = f0(g(x))g0(x)Solution: True. Chain rule.(e) True or False. A function can have two different horizontal asymptotes.Solution: True. limx→−∞f(x) does not have to equal limx→∞f(x) such as when f(x) =arctan x.(f) True or False. limx→42xx − 4−8x − 4= limx→42xx − 4− limx→48x − 4.Solution: False. The limit of a difference is the difference of the limits only ifboth limits exist.12. Complete the following sentence:The function f(x) is continuous on the interval [a, b] ifSolution: limx→cf(x) = f(c) for every number c in the interval (a, b).3. Compute the following limits. Justify your results. If the limit is positive or negativeinfinity, it should be clearly indicated instead of just saying the limit does not exist(a) limx→πcos(x + sin(x))Solution:limx→πcos(x + sin(x)) = cos(π + sin(π)) b/c continous at x = π.= cos(π) = 1(b) limx→12 − x(x − 1)2Solution:limx→12 − x(x − 1)2= ∞because the numerator, 2 − x approaches 1 as x → 1 while the denominatorbecomes a very small positive number as x → 1. Thus the fraction becomes avery large positive number as x → 1.(c) limx→−414+1x4 + xSolution:limx→−414+1x4 + x= limx→−4x+44x4+x1= limx→−414xb/c limit doesn’t see x = −4=−116b/c continuous at x = −4.2(d) limx→−∞2x2− 5x − 2x4− 10x2− 3Solution:limx→−∞2x2− 5x − 2x4− 10x2− 3= limx→−∞2x2x4−5xx4−2x4x4x4−10x2x4−3x4= limx→−∞2x2−5x3−2x41 −10x2−3x4=limx→−∞2x2− limx→−∞5x3− limx→−∞2x41 − limx→−∞10x2− limx→−∞3x4=01= 04. Use the limit definition of the derivative to compute the derivative of f(x) =√1 + 2x.Solution:f0(x) = limh→0f(x + h) −f(x)h= limh→0p1 + 2(x + h) −√1 + 2xh= limh→0(p1 + 2(x + h) −√1 + 2x)h(p1 + 2(x + h) +√1 + 2x)(p1 + 2(x + h) +√1 + 2x)= limh→01 + 2(x + h) − (1 + 2x)h(p1 + 2(x + h) +√1 + 2x)= limh→02hh(p1 + 2(x + h) +√1 + 2x)= limh→02(p1 + 2(x + h) +√1 + 2x)b/c limit doesn’t see h = 0=2(p1 + 2(x + 0) +√1 + 2x)b/c continuous at h = 0=1√1 + 2x35. Compute the following derivatives. You do not have to use the definition of the deriva-tive. If you can “do them in your head” instead of showing every step that is up toyou (though if you get it wrong we cannot give you partial credit.)(a) Let f(x) = 3x ln(x). Find f0(x).Solution:f0(x) = ln(3)3x ln(x)(ln(x) +xx)= ln(3)3x ln(x)(ln(x) + 1)(b) Findddxpx ln(x4).Solution:ddxpx ln(x4) =12(x ln(x4))−1/2(ln(x4) + x(1x4(4x3)))=12(x ln(x4))−1/2(ln(x4) + 4)(c) Findddx(ln(x))cos(x).Solution: We must use logarithmic differentiation. So lety = (ln(x))cos(x)ln(y) = ln((ln(x))cos(x))= cos(x) ln(ln(x))1ydydx= −sin(x) ln(ln(x)) + cos(x)(1ln(x)1x)= −sin(x) ln(ln(x)) + cos(x)(1x ln(x))dydx= y(−sin(x) ln(ln(x)) + cos(x)(1x ln(x)))= (ln(x))cos(x)(−sin(x) ln(ln(x)) + cos(x)(1x ln(x)))4(d) Findddx(4x − 1)3(2x2− 1)3/2(x + 1)2.Solution: We can use logarithmic differentiation to make this easier. So lety =(4x − 1)3(2x2− 1)3/2(x + 1)2ln(y) = ln(4x − 1)3(2x2− 1)3/2(x + 1)2= ln((4x − 1)3) − ln((2x2− 1)3/2) − ln((x + 1)2)= 3 ln(4x − 1) −32ln(2x2− 1) − 2 ln(x + 1)1ydydx= 314x − 1(4) −3212x2− 1(4x) − 21x + 1=124x − 1−6x2x2− 1−2x + 1dydx= ((4x − 1)3(2x2− 1)3/2(x + 1)2)(124x − 1−6x2x2− 1−2x + 1)(e) Findddx(x2cos(x)e3xsin(π/2)).Solution:ddx(x2cos(x)e3xsin(π/2)) =ddx(x2cos(x)e3x)= 2x cos(x)e3x− x2sin(x)e3x+ x2cos(x)(3e3x)56. Find the equation of the line tangent toy = (2 + x)e−xat the point (0, 2)Solution: First we must finddydx.dydx= e−x+ (2 + x)e−x(−1)= e−x(1 − (2 + x))So the slope of the tangent line to this curve at (0, 2) is e0(1 −(2 −0)) = 1(−1) = −1.Therefore the equation of the line tangent to this curve at (0, 2) is y −2 = (−1)(x).7. Show that there exists a solution to the equation ln(x) = sin(π2x) on the interval (0, ∞).Solution: To prove that there is a solution to this equation on the given interval, weneed to use the Intermediate Value Theorem. First we need to have a function solet f(x) = ln(x) − sin(π2x). We need to know that this function is continuous on theinterval and that it there are numbers a and b in (0, ∞) so that f (a) < 0 and f(b) > 0.Notice that both ln(x) and sin(π2x) are continuous functions on (0, ∞) so f (x) is aswell. Notice also thatf(1) = ln(1) − sin(π2) = 0 − sin(π/2) = −1 < 0.Finally,f(e5) = ln(e5) − sin(π2e5) = 5 − sin(π2e5) ≥ 5 − 1 = 4 > 0since sin(anything) ≤ 1. Therefore by the intermediate value theorem, there is a numberc between 1 and e5so that ln(c) = sin(π2c) as desired.68. Consider the function f (x) =11 − x2.(a) Write the domain of f(x) in interval notation and find the coordinates of allpoints, if any, where the graph of f crosses the x-axis.Solution: Notice that f(x) is defined for every real number except when 1−x2= 0,in other words, except when x = −1, 1. Thus the domain of f(x) is (−∞, −1) ∪(−1, 1) ∪ (1, ∞).Finally, f(x) crosses the x−axis when f(x) = 0 so when11−x2= 0 which happensonly when 1 = 0 - which never happens! Thus f(x) never crosses the x−axis.(b) Find the equations for all of the vertical and horizontal asymptotes of this func-tion, or state why there are none. Justify asymptotes with limit calculations.Solution: We see that f (x) is defined and continuous everywhere except at x = −1and x = 1. Then the only possible candidates for vertical asymptotes are atx = −1 and x = 1.When x = −1, consider limx→−1−f(x). The numerator is always 1 and the de-nominator is a small negative number so f(x) is a very large