Math 231E. Fall 2013. Midterm 1 Practice Solutions.Problem 1. Compute the derivatives of the following functionsa. f(x) = sin(x2)exb. f(x) = arcsec(x)c. f(x) = ln(1 + x2+ x4)Solution:a. We use the product and Chain Rules:ddx(sin(x2)ex) =ddx(sin(x2))ex+ sin(x2)ddxex= 2x cos(x2)ex+ exsin(x2).b. We have to use implicit methods here. We write y = sec−1(x), orx = sec(y).Differentiating both sides gives1 = sec(y) tan(y)dydx=sin(y)cos2(y)dydx,sodydx=cos2(y)sin(y).Draw a triangle with near side of length x and far side of length 1, and if the interiorangle is y, then sec(y) = x. Since the hypotenuse of the triangle is√1 + x2, this meansthat1√x2− 1xsin(y) =√x2− 1x, cos(y) =1x,sodydx=1x√x2− 1.c. Using the Chain Rule,f0(x) =2x + 4x31 + x2+ x4.Problem 2. Give the Taylor polynomial for the following functions to the order indicatedabout the point a = 0a. f(x) = (sin(x) − x)ex, order = 4b. f(x) = (cos(x))/(1 − x), order = 3c. f(x) =11 + x2+ x4, order = 8d. f(x) = x(cos(x2) − 1) sin(x), order = 10e. f(x) = ln(1 + x2), order 6Solution:a. We write out each one as follows:sin(x) − x = −x36+ O(x5),ex= 1 + x +x22+x36+ O(x4).Multiplying these gives(sin(x) − x)ex=−x36+ O(x5)1 + x +x22+x36+ O(x4)= −x36−x46+ O(x5).b. We could do long division /. But we can be more clever. Recall that1x − 1= 1 + x + x2+ x3+ O(x4).We also knowcos(x) = 1 −x22+ O(x4).Thuscos(x)1 − x=1 + x + x2+ x3+ O(x4)1 −x22+ O(x4)=1 + x + x2+ x3+ O(x4)−x221 + x + x2+ x3+ O(x4)+ O(x4)= 1 + x + x2−x22+ x3−x32+ O(x4)= 1 + x +x22+x32+ O(x4).c. We could again do long division, and again /. But let us again use the fact that11 − x= 1 + x + x2+ x3+ x4+ O(x5),and thus11 + x=11 − (−x)= 1 + (−x) + (−x)2+ (−x)3+ (−x)4+ O(x5)= 1 − x + x2− x3+ x4+ O(x5).Finally, let us plug in x2+ x4for x in the last series, so11 + x2+ x4= 1 − (x2+ x4) + (x2+ x4)2− (x2+ x4)3+ (x2+ x4)4+ O(x10).We know that(x2+ x4)2= x4+ 2x6+ x8,(x2+ x4)3= x6+ 3x8+ O(x10),(x2+ x4)4= x8+ O(x10),and thus we have11 + x2+ x4= 1 − (x2+ x4) + (x4+ 2x6+ x8) − (x6+ 3x8+ O(x10)) + (x8+ O(x10)) + O(x10)= 1 − x2− x4+ x4+ 2x6+ x8− x6− 3x8+ x8+ O(x10)= 1 − x2+ x6− x8+ O(x10).d. Products! No worries:cos(x) − 1 = −x22+x424+ O(x6),cos(x2) − 1 = −x42+x824+ O(x12),sin(x) = x −x36+x5120+ O(x7),sox(cos(x2) − 1) sin(x) = x−x42+x824+ O(x12)x −x36+x5120+ O(x7)= x−x42x −x36+x5120+ O(x7)+x824x −x36+x5120+ O(x7)+ O(x12)= x−x52+x712−x9240+ O(x11) +x924+ O(x11)+ O(x12)= −x62+x812−x10240+x1024+ O(x12)= −x62+x812+3x1080+ O(x11).e. We could do this directly, but it will be easier to do a Taylor series for ln(1 + x) andthen plug in. So let us say g(x) = ln(1 + x), theng0(x) =11 + x, g00(x) =−1(1 + x)2, g000(x) =2(1 + x)3,and thusln(1 + x) = 0 + x −x22+x33+ O(x4),and thusln(1 + x2) = x2−x42+x63+ O(x8).Problem 3. Give the Taylor series for the following functions to third order about the pointindicateda. f(x) = cos(x), about the point a =π2b. f(x) = sin(x), about the point a =π4c. f(x) =√x, about the point a = 2Solution:a. We compute thatf0(x) = −sin(x), f00(x) = −cos(x), f000(x) = sin(x),and thusf(π/2) = 0, f0(π/2) = −1, f00(π/2) = 0, f000(π/2) = 1,sof(x) = −(x − π/2) +16(x − π/2)3+ O((x − π/2)4).b. We compute thatf0(x) = cos(x), f00(x) = −sin(x), f000(x) = −cos(x),and thusf(π/4) =√22, f0(π/2) =√22, f00(π/2) ==√22, f000(π/2) = −√22,sof(x) =√22+√22(x − π/4) −√24(x − π/4)2−√212(x − π/4)3+ O(x − π/4)4.c. We havef0(x) =12√x, f00(x) =−14x3/2, f000(x) =38x5/2,sof(2) =√2, f0(2) =12√2, f00(2) = −18√2, f000(2) =332√2,sof(x) =√2 +12√2(x − 2) −116√2(x − 2)2+164√2(x − 2)3+ O(x − 2)4.Problem 4. Some triangle problems!a. Suppose u = cos(x). Express sin(x) in terms of u.b. Given that u = tan(x), express sin(x) in terms of u.Solution:a. We draw a triangle so that cos(x) = u, obtainingu√1 − u21The Pythagorean theorem tells us that the far side is√1 − u2, so sin(x) =√1 − u2.b. We draw a triangle so that u = tan(x), obtaining1u√1 + u2Thussin(x) =u√1 + u2.Problem 5. Consider the polynomial P (x) = 6 −14x −3x2+ 5x3. Let A, B be the intervalsdefined as:A = {x : −1 < x < 0}, B = {x : 0 < x < 1}Using the Intermediate Value Theorem, in which intervals is the polynomial P (x) guaranteedto have a zero? Explain why.a. Both A and Bb. Only Ac. Only Bd. Neither A nor Be. Not enough informa-tion to decide.Solution: We compute thatP (−1) = 12, P (0) = 6, P (1) = −6.From this and the Intermediate Value Theorem, we know that there must be a root between0 and 1, since 0 lies between 6 and −6.Problem 6. Evaluate the following limitsa. limx→11 − 3x2+ 2x3cos(x)b. limx→0|x|cos(x)c. limx→0sin(x)xd. limx→0cos(x2) − 1x4e. limx→0ex3− 1x − sin(x)f. limx→11 + x + x2(x − 1)2g. limx→2sin(πx)ex−2− 1h. limx→∞sin(x2) + 11xx2+ exi. limx→∞x14+ 11x7− 9727x14+ 19x13− 293x5− 5j. limx→0|x|xk. limx→0x2ecos(x)sin(x)Solution:a. Continuous, so substitute: f (1) =1 − 3 + 2cos(1)= 0b.limx→0+|x|cos(x)= limx→0+xcos(x)= 0andlimx→0−|x|cos(x)= limx→0−−xcos(x)= 0Thuslimx→0+|x|cos(x)= 0c. Taylor serieslimx→0sin(x)x= limx→0x −x36+ O(x5)x= limx→01 −x26+ O(x4) = 1or L’Hopitald. Taylor serieslimx→0cos(x2) − 1x4= limx→0−x4/2 +x824+ O(x12)x4= limx→0−1/2 + O(x4) = −1/2(L’Hopital = /)e. Taylor Serieslimx→0ex3− 1x − sin(x)= limx→0x3+ O(x6)x3/6 + O(x5)= limx→06 + O(x2) = 6f. Denominator goes to zero, is positive, numerator goes to 3, thuslimx→11 + x + x2(1 − x)2= ∞g. L’Hopital! limx→2sin(πx)ex−2− 1= limx→2π cos(πx)ex−2= π cos(2π) = πh. Squeeze11x − 1x2+ ex<sin(x2) + 11xx2+ ex<11x + 1x2+ exand both limit on left and on right are 0 (L’Hopital twice, or use fact that exgrowsfaster than any power.i. limx→∞x14+ 11x7− 9727x14+ 19x13− 293x5− 5= limx→∞1 + 11x−7− 972x−147 + 19x−1− 293x−9− 5x−14=17j. Does not exist. limx→0+|x|x= 1 and limx→0−|x|x= −1k. Squeeze + Taylorx2e−1sin(x)<x2ecos(x)sin(x)<x2e1sin(x)And the left and the right limit are both zero by L’Hopital or by Taylor series.Problem 7.f(x) = “the seventh digit after the decimal point in the decimal expansion for x”.DefineA = limx→1−f(x), B = limx→1+f(x).Compute A, B.Solution: If x is very close to 1 (say within 10−8) and is above 1, then
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