PRACTICE PROBLEMS FOR MIDTERM – SOLUTIONSThe test will be given onWednesday, April 28.It will cover Sections17-20 and 28-29 (everything covered in this quarter, up to and includingMean Value Theorem).In preparing for the test, you can practice solving the problems from the listbelow. In addition, take a look at the homework assignments 1-3 (at leastone problem on the midterm will come from the homework), at the examplesgiven in the textbook, and the the examples discussed in class.1.Suppose the functionsf,g, andhare defined on the interval (a, b), andg(x)≤f(x)≤h(x) for everyx∈(a, b). Suppose, furthermore, thatgandhare continuous atc∈(a, b), andg(c) =h(c). Prove thatfis continuousatc.We need to show that, if (xn) is a sequence of elements of (a, b), satis-fying limxn=c, then limf(xn) =f(c). Butg(xn)≤f(xn)≤h(xn), andlimg(xn) =g(c) =f(c), and limh(xn) =h(c) =f(c). By Squeeze Theorem,limf(xn) =f(c).2.Prove that there existsx∈[0,2] s.t.x5−3x−1 =√x3+ 1.Considerf(x) =x5−3x−1−√x3+ 1.Thenfis continuous on [0,2].Indeed,f(x) =g(x)−h(x), whereg(x) =x5−3x−1,h(x) =radicalbigk(x),andk(x) =x3+ 1.gandkare polynomials, hence they are continuous.Moreover,k(x)≥0 for anyx∈[0,2], hencehis also continuous on [0,2].This implies the continuity off=g−h.Observe thatf(0) =−2, andf(2) = 32−6−1−3 = 22. By Intermediatevalue Theorem,f(x) = 0 for somex∈(0,2).Therefore,x5−3x−1 =√x3+ 1.3.Letf(x) = 1/√x. Prove thatfis not uniformly continuous on (0,1].Solution 1. We know that a uniformly continuous function maps a Cauchysequence to a Cauchy sequence. This is not the case forf. Indeed, letsn=1/n(n∈N).The sequence (sn) is convergent, hence Cauchy.Moreover,sn∈(0,1] for anyn. However,f(sn) =√n, and limf(sn) = +∞. Thus,the sequence (f(sn)) is not Cauchy.