Qz5RevM3310S16 - Math 3310 Theoretical Concepts of Calculus Quiz 5 Review Section 5.3 We will use the intermediate value theorem in the following

Qz5RevM3310S16 - Math 3310 Theoretical Concepts of Calculus...

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Math 3310Theoretical Concepts of CalculusQuiz 5 ReviewSection 5.3We will use the intermediate value theorem in the following problems.Exercises4 :Show that3x= 5xfor somex2(0;1).Solution:Letf(x) = 3x°5x:Notice3x=5xhas a solution inside(0;1)if and only iff(x)=0forx2(0;1):Sincefis continuous onR, andf(0) = 1andf(1) = 3°5 =°2, by the IVT, we have that9c2(0;1)(f(c) = 0):Thus, there isc2(0;1);such that3c= 5c:Exercise5 :Show that5x=x4has at least one real solution.Solution:Letf(x) = 5x°x4:Notice5x=x4has at least one real solution if and only iff(x)=0for somex2R.Sincefis continuous onR, andf(°1) =15°1 =°45andf(0) = 1°0 = 1, by the Intermediate ValueTheorem, we have that9c2(°1;0)(f(c) = 0):Thus, there isc2(°1;0);such that5c=c4:Therefore,5x=x4has at least one real solution.Exercise6 :Show that any polynomial of odd degree has at least one real root.Solution:Letf(x) =anxn+an°1xn°1+:::+a1x+a0; ai2R; i= 0;1; :::; n;be polynomial of degreen >0;andnis odd. Assume thatan>0:Thenlimx!°1f(x) =°1andlimx!1f(x) =1:It follows, that8M >09a;b2Rf(a)<°Mandf(b)> MThus, in particular, we havef(a)<0andf(b)>0Since all polynomial functions are continuous, then by the IVT, we have that9c2(a; b)(f(c) = 0):Thus, there isc2(a; b);such thatf(c) = 0;that is,fhas at least one real root:Section 5.4Exercise3: Determine which of the following continuous functions are uniformly continuous on thegiven set. Justify your answer.(a)f(x) =exxon[2;5]Solution:Since the set[2;5]±Ris closed and bounded, thus it is compact by Heine-Borel theorem.Since,fis continuous onR±(as quotient of two continuous functions), it follows thatfis also uniformlycontinuous on[2;5].1
(b)f(x) =exxon(0;2)Solution:We observe thatlimx!0+exx=1:Therefore, the functionfcannot be extended to a functionefthat is continuous on[0;2]. It follows thatfis not uniformly continuous on(0;2).(c)f(x) =x2+ 2x°7on[0;5]Solution:Since the set[0;5]±Ris closed and bounded, thus it is compact by Heine-Borel theorem.Since,fis continuous onR(fis a polynomial function), it follows from thatfis also uniformly continuouson[0;5].(d)f(x) =x2+ 2x°7on(1;4)Solution:We observe thatlimx!1+°x2+ 2x°7±=°4andlimx!4°°x2+ 2x°7±= 17:Therefore, the functionfcan be extended to a functionefthat is continuous on[0;4];namely, we de°ne:ef:[0;4]!Ref(x)=x2+ 2x°7It follows thatfis uniformly continuous on(1;4).(e)f(x) =1x2on(0;1)Solution:We observe thatlimx!0+1x2=1:Therefore, the functionfcannot be extended to a functionefthat is continuous on[0;1]. It follows thatfis not uniformly continuous on(0;1).(h)f(x) =xsin°1x±on(0;1)Solution:We observe thatlimx!0xsin²1x³= 0Therefore, the functionfcan be extended to a functionefthat is continuous on[0;1];namely, we de°ne:ef:[0;1]!Ref(x)=´xsin°1x±ifx2(0;1]0ifx= 0It follows thatfis uniformly continuous on(0;1).

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