Week 1: Calculus I Practice Problem Solutions

1 Christian ParkinsonGRE Prep: Calculus I Practice Problem Solutions1 Week 1: Calculus IPractice Problem SolutionsProblem is the tangent line to the graph ofy=x+exatx= 0? tangent line is given by`(x) =y(0) +y (0)(x 0) = 1 + limx 0(1 +x) 1xfor l H opital s rule, we seelimx 0(1 +x) 1x= limx 0 (1 +x) 11= .This gives a first order approximation (1 +x) 1 + xwhenx limx 0cos( x) 1x2for l H opital s rule, we seelimx 0cos( x) 1x2= limx 0 sin( x)2x= limx 0 2cos( x)2= >0. Find the minimum value off(x) =ex cxamongx the derivative to zero shows that extreme points occur whenex c= 0 x= log(c).The second derivative offis always positive so any extreme point is a minimum.

2 Thus theminimum value isf(log(c)) =c clog(c). Problem (x) =|x|+ 3x2forx R. What isf ( 1)? a neighborhood of 1, we have|x|= xand sof(x) = x+ 3x2. Thenf ( 1) = 1 + 6( 1) = the limit limx 0(x 2 sin(x) 2).Christian ParkinsonGRE Prep: Calculus I Practice Problem seelimx 0(x 2 sin(x) 2) = limx 0sin2(x) x2x2sin2(x)= limx 0 (x2sin2(x)) 1(sin2(x) x2x4) = limx 0 (sin(x) +xx) 2(sin(x) xx3) = 2 limx cos(x) 13x2= 2 limx 0 sin(x)6x= thatf(x) = 3x2+bx+chas a non-simple root atx= 2. What isf(5)? a quadratic polynomial has a non-simple root atx= 2 then it is a multiple of(x 2)2. Heref(x) = 3(x 2)2= f(5) = 3(3)2= :R Ris continuously differentiable on ( 1,4) withf(3) = 5 andf (x) 1 for allx ( 1,4), what is the greatest possible value off(0)?

3 The fundamental theorem of Calculus , we havef(0) =f(3) 30f (x)dx f(3) 301dx= bound can be realized iff (x) 1 [sof(x) =x+ 2]. Problem limx 0sin(2x)(1 +x) ln(1 +x). term11+xis irrelevant since it tends to 1 in the limit. Thuslimx 0sin(2x)(1 +x) ln(1 +x)= limx 0sin(2x)ln(1 +x)= limx 02(1 +x) cos(2x) = (x) =eg(x)h(x) whereh (x) = g (x)h(x) for allx R. Which of thefollowing is necessarily true? (a)fis constant(b)fis linear and non-constant(c)gis constant(d)gis linear and non-constant(e) none of the thatf (x) =eg(x)g (x)h(x) +eg(x)h (x) =eg(x)g (x)h(x) eg(x)g (x)h(x) = 0 Christian ParkinsonGRE Prep: Calculus I Practice Problem Solutions3sofis (x) =x2+sin(x)forx >0.

4 Findf (x). temptation here is to use the power rule or the exponential rule but in thecurrent form, neither apply since both the base and the exponent depend onx. To fix this,we writef(x) =e(2+sin(x)) log(x). Thusf (x) =e(2+sin(x)) log(x)(2 + sin(x)x+ cos(x) log(x))=x2+sin(x)(2 + sin(x)x+ cos(x) log(x)). Problem 10 1 x4dx, K= 10 1 +x4dx, L= 10 1 thenumbersJ,K,L,1 in increasing (0,1), we have1 x4<1 x8<1<1 + the square root and integrating shows thatJ < L <1< Rsuch thatg:R Rsatisfies 3x5+ 96 = xcg(t) shows thatg(x) = 15x4. Thus3x5+ 96 = 3x5 3c5= c= (0) = 0 andf(x) =|x|xforx6= 0. Compute 1 1f(x) function is odd so the integral on a symmetric range is (x) = x1dt1+t2.

5 Find an equation for the tangent line at (2,f(2)). tangent line is given by`(x) =f(2) +f (2)(x 2).Heref(2) = arctan(2) 4andf (2) =11+22=15so`(x) = arctan(2) 4+15(x 2). Problem (x) = x0cos2(t2)dt. Find (f 1) (y) fory=f(3).Christian ParkinsonGRE Prep: Calculus I Practice Problem that (f 1) (y) =1f (f 1(y)). Thus(f 1) (f(3)) =1f (3)= sec2(9). Problem continuous functionsf,g:R R, define the relation byf gifflimx f(x)g(x)= thatf g. Which of these does NOT necessarily follow:(a)f2 g2(b) f g(c)ef eg(d)f+g 2g(e)g one which does NOT follow is (c)ef eg. Indeed, putf(x) =xandg(x) =x 1. Thenf g, butlimx ef(x)eg(x)= limx e=e6= 1soef6 (x) =e2x+1.

6 Compute limx 0g(g(x)) g(e) key is to recognize the limit as (g g) (0). Now(g g) (x) =ddx(e2e2x+1+1)=e2e2x+1+1(4e2x+1)so (g g) (0) =e2e+1 (4e) = 4e2e+ thatfis differentiable atx=x0. What is limh 0f(x0+h) f(x0 h)h? adding and subtractingf(x0) in the middle of the numerator, we see thatthis limit is 2f (x0). Problem the derivativeddx x20e the fundamental theorem of Calculus and the chain ruleddx x20e t2dt= 2xe the first derivative off(x) =x3(6x2+1)3 (x+3)4whenx > ParkinsonGRE Prep: Calculus I Practice Problem can vastly simplify the Problem using logarithmic differentiation. Indeed,log(f(x)) = 3 log(x) log(6x2+ 1) 43log(x+ 3).Thusf (x)f(x)=3x 12x6x2+ 1 43(x+ 3)andf (x) =x3(6x2+ 1)3 (x+ 3)4(3x 12x6x2+ 1 43(x+ 3)).

7 Problem limn 2n k=n+ 2n k=n+11k= limn n k=11n+k= limn 1nn k=111 + is a limit of Riemann sums for11+xon [0,1]. Thuslimn 2n k=n+11k= 10dx1 +x= ln(2).Note: this actually proves that the harmonic series k=11kdiverges. Indeed, looking atthe partial sumsHn= nk=11k, we have proven that limn (H2n Hn) = ln(2). But thenfor allmsufficiently large, we haveH2m Hm>1/2 which shows that the sequence{Hn}is not a Cauchy many real roots does 2x5+ 8x 7 have? the polynomial has odd order it has at least one real root (by the interme-diate value theorem). The derivative of the polynomial is 10x4+ 8 which is always positiveso the polynomial is always strictly increasing and thus has at most one limx 0{1x x0(1 + sin(2t))1/tdt}.

8 (x) = x0(1 + sin(2t))1/tdt, we see that the Problem is asking forF (0).By FToC, we haveF (x) = (1 + sin(2x))1 (0) = limx 0(1 + sin(2x))1/x= exp(limx 0log(1 + sin(2x))x)= exp(limx 02 cos(2x)/(1 + sin(2x))1)= ParkinsonGRE Prep: Calculus I Practice Problem Solutions6 Note: to be more rigorous, you would actually need to show thatF (0) exists and is equal tothis limit; on the GRE you can dispose of theoretical concerns like this for the sake of time,and because you will be given options for the : [0,1] Rbe continuous and suppose thatfis differentiable on (0,1)withf(0) = 1,f(1) = 0. Which of the following are necessarily true?(a) There isx (0,1) such thatf(x) =x(b) There isx (0,1) such thatf (x) = 1(c)f(x)>0 for allx [0,1)Solution.]

9 (a) is true by applying the intermediate value theorem tof(x) x. (b) is trueby the mean value theorem. (c) is not necessarily true (as can be easily seen by drawing apicture). Problem 3 3|x+ 1| are several ways to do this. One way is to explicitly calculate the integralby splitting up the region; another is to notice that we are simply adding the areas of twoisosceles right trangles. The answer is limn calculating the integral giveslimn n1dxxn= limn n1 n 11 n= , you could use the dominated convergence theorem to see that this is zerowithout the annoying limn 3nn i=1[(3in)2 (3in)]. can recognize this as a limit of Riemann sums forx2 xwith step size 3/non [0,3] or as thrice the limit of Riemann sums of (3x)2 3xwith step size 1/non [0,1].

10 Thuslimn 3nn i=1[(3in)2 (3in)]= 30(x2 x)dx= 9 9/2 = 9/2orlimn 3nn i=1[(3in)2 (3in)]= 3 10(9x2 3x)dx= 3(3 3/2) = 9/2.