EE 518 : Homework #5 Due on Monday 2:00pm, October 3, 2016Problem 1Suppose that f : (0, 1) → R is uniformly continuous on (0, 1). If {an} is a Cauchy sequence in (0, 1) andbn= f(an), show that {bn} is a Cauchy sequence in R.Let  > 0 be an arbitrary real number. Since f is uniformly continuous on (0, 1), then there exists a δ > 0such thatif |x −y| < δ and x, y ∈ (0, 1) ⇒ |f(x) − f(y)| < .Since {an} is a Cauchy sequence in (0, 1), then for every δ > 0 there is an integer N such that |an− am| < δif n, m > N. Then from the uniform continuity of f, it follows that |f(an) − f(am)| < , if n, m > N,which implies that {f(an)} is a Cauchy sequence in R.Note: The above result is not true if f is only continuous. For example, f(x) =1xis continuous on (0, 1)and let an=1n, then bn= f(an) = n is not a Cauchy sequence.Problem 2True or false? Justify your answer. Give a counterexample if it is false.(a) If limx→af(x) = 0 and limx→ag(x) = 0, then limx→af(x)g (x)does not exist.(b) If neither limx→af(x) nor limx→ag(x) exists, then limx→a(f(x) + g(x)) does not exist.(c) If limx→af(x) exists, but limx→ag(x) does not exist, then limx→a(f(x) + g(x)) does not exist.(d) If limx→a(f(x)g(x)) exists, then the limit must be equal to f(a)g(a).(e) If limx→af(x) = ∞ and limx→ag(x) = ∞, then limx→a(f(x) − g(x)) = 0.(a) False. Counterexample: limx→1x(x−1)x−1or limx→1sin(x−1)x−1.(b) False. Counterexample: Let f(x) =1xand g(x) =−1x, then limx→0f(x) and limx→0g(x) do not exist,but limx→0(f(x) + g(x)) = limx→00 = 0 exists.(c) True. By contradiction, assume limx→a(f(x) + g(x)) exists, then using the properties of limit wecan write limx→ag(x) = limx→a(f(x) + g(x) − f(x)) = limx→a(f(x) + g(x)) − limx→af(x). Sincelimx→a(f(x) + g(x)) and limx→af(x) both exist, then limx→ag(x) exists! This contradiction shows thatour first assumption that limx→a(f(x) + g(x)) exists is wrong.(d) False. Counterexample: Let f(x) = x − 1 and g(x) =1x−1, then limx→1(f(x)g(x)) = 1 exists, butf(1) = 0 and g(1) is undefined.(e) False. Counterexample: Let f(x) =1x2and g(x) =1x4, then limx→0f(x) = limx→0g(x) = ∞, butlimx→0(f(x) − g(x)) = limx→0x2−1x4= −∞.Problem 3Find any local minimum or local maximum of following functions, if exist.(i)f(x) =12x2e−x(ii)f(x) =x1 + x2Page 1 of 4EE 518 : Homework #5 Due on Monday 2:00pm, October 3, 2016(i) Since f (x) is well-defined and differentiable on R, we only consider when f0(x) = 0. Letf0(x) = xe−x−12x2e−x= 0we have 2 solutionsx1= 0, x2= 2.Sincef00(x) = (1 − 2x +12x2e−x)f00(0) = 1 > 0, f00(2) = −e−2< 0Therefore, x1= 0 is a local minimum and x2= 2 is a local maximum.(ii) Letf0(x) =1 − x2(1 + x2)2= 0We havex1= −1, x2= 1since f(−1) = −1/2 < f (1) = 1/2, therefore x1= −1 is a local minimum and x2= 1 is a localmaximum.This only works for continuous functions and only two local optimum exists.Problem 4(a) A critical point of a function f : X → R is a point c ∈ X such that either f0(c) = 0 or f0(c) does notexist. Consider the function f(x) = ax3+ bx2+ cx + d, where a 6= 0. Show that f can have two, one,or no critical points (give examples and sketches to illustrate the three possibilities). How many localextreme values can f have?(b) Show that the equation ex− 2 = cos√x has at least one root in (0, 1).(c) Consider the function f (x) = x4− 4x3+ 4x2+ c, where c is a constant. Show that f has at most onezero in the interval (1, 2). For what interval of values of c does the equation f(x) = 0 have exactly oneroot in (1, 2)?(a) f0(x) = 3ax2+ 2bx + c is a 2nd-order polynomial, so it has either two, one, or zero real roots. Thus, fcan have either two, one, or zero critical points. For example, (i) f(x) = x3− 3x has two critical points,(ii) f(x) = x3has one critical point, and (iii) f(x) = x3+ 3x has no critical point.It can have at most two local extreme values (case (i)). For cases (ii) and (iii), there is no local extremevalue. See Fig.1.(b) The function f(x) = ex− 2 − cos√x is continuous on [0, 1]. f(0) = −2 and f(1) ≈ 0.178. Sincef(0) < 0 < f(1), then from the Intermediate Value Theorem, there exists a c ∈ (0, 1) such that f(c) = 0.Then,, ex− 2 = cos√x has at least one root in (0, 1).(c) f0(x) = 4x3− 12x2+ 8x = 4x(x2− 3x + 2) = 4x(x − 1)(x − 2) < 0 on (1, 2). Since f is strictlydecreasing on (1, 2), then if there is any root, it is unique. So, f has at most one zero in the interval(1, 2). Since f is continuous and strictly decreasing on (1, 2), and f(1) = 1 + c and f (2) = c, then fromthe Intermediate Value Theorem, f(x) = 0 has exactly one root if and only if f(1) and f(2) have oppositesigns, i.e., c < 0 < c + 1, or −1 < c < 0.Page 2 of 4EE 518 : Homework #5 Due on Monday 2:00pm, October 3, 2016Figure 1: Sketch for Problem 4aProblem 5Suppose function f has n derivatives on the interval (0, a) and limx→0+x2nf(n)(x) = L exists in R, wheref(n)denotes the n-th derivative of f. Find limx→0+xnf(x) in terms of L.Apply L’Hospital’s rule n times in succession to f(x)/x−nwe obtainlimx→0+xnf(x) = limx→0+(−1)nf(n)(x)n(n + 1) . . . (2n −1)x−2n=(−1)nn(n + 1) . . . (2n −1)limx→0+x2nf(n)(x)=(−1)nn(n + 1) . . . (2n −1)L =(−1)n(n − 1)!(2n − 1)!L.Problem 6Given values of variables x and y:x = [0 0.3 0.8 1.1 1.6 2.2];y = [0.6 0.67 1.01 1.35 1.45 1.20];Fit the data with first, second, and fifth degree polynomials and plot all polynomial fits with different colorson the same figure. Submit your code along with your plot. Hint: use polyfit() and polyval()Page 3 of 4EE 518 : Homework #5 Due on Monday 2:00pm, October 3, 2016Page 4 of