Math 242: Principles of Analysis Fall 2016 Homework 7 Part B Solutions

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1 Mat 22: Principles of Analysis Fall 206 Homework 7 Part B Solutions. Sow tat f(x) = x 2 is not uniformly continuous on R. Solution. Te equation is equivalent to f(x) = 0 were f(x) = x 2 sin(x) 3. Since f(0) = 3 < 0 an f(3) = 6 sin(3) > 5 > 0 an f is continuous on [0, 3], te Intermeiate Value Teorem implies tat tere exists c (0, 3) suc tat f(c ) Since f( 3) = 6 sin( 3) > 5 > 0 an f is continuous on [ 3, 0], te Intermeiate Value Teorem implies tat tere exists c 2 ( 3, 0) suc tat f(c 2 ) Since c > 0 an c 2 < 0, tese roots are istinct. 2. A function f is Lipscitz on an interval I if tere exists a constant L > 0 suc tat f(y) f(x) L y x for every x, y I. (a) Sow f(x) = x 2 is Lipscitz on [ M, M] for any M > 0. Solution. For any x, y [ M, M] we ave f(y) f(x) = y 2 x 2 = y + x y x ( y + x ) y x 2M y x, an tus f is Lipscitz on [ M, M] wit constant L = 2M. (b) Sow tat if f is Lipscitz on I, ten f is uniformly continuous on I. Solution. Given ϵ > 0, coose δ = ϵ/l. Ten for any x, y I suc tat y x < δ we ave f(y) f(x) L y x < Lδ = ϵ. Tus f is uniformly continuous on I. (c) Sow f(x) = x is uniformly continuous on [0, ], but not Lipscitz on [0, ]. Solution. Since f is continuous on [0, ], f is uniformly continuous on [0, ]. Suppose f were Lipscitz on [0, ]. Ten for some constant L > 0 we woul ave f(y) f(x) L y x for all x, y [0, ]. Fixing x = 0, tis woul imply f(y) L y for all y [0, ]. Setting y = for eac n N ten gives n f(/n) L/n. Tis is equivalent to n L/n, or n L for all n N. Since tere is no suc constant L, f is not Lipscitz. () Prove tat if f as erivative f tat is continuous on [a, b], ten f is Lipscitz on [a, b]. Solution. Since f is continuous on [a, b], te Extreme Value Teorem implies tat f attains a maximum an a minimum on [a, b]. Tus f is boune on [a, b], so tere exists some M > 0 suc tat f (x) M for all x [a, b]. Given any x, y [a, b], we may suppose witout loss of generality tat x < y. Ten applying te Mean Value Teorem to f on [x, y], tere exists some c (x, y) suc tat f(y) f(x) = f (c)(y x). Hence so f is Lipscitz on [a, b]. f(y) f(x) = f (c) y x M y x

2 3. Use te efinition to fin te erivative of eac function. (a) f(x) = x Solution. For c 0 we ave Tus f (x) = x 2. f (c) x c (b) s(x) = sin(x) Solution. For any c R we ave Tus s (x) = cos(x). x c x c x c c x xc x c x c xc = c. 2 s sin(c + ) sin(c) (c) sin(c) cos() + sin() cos(c) sin(c) sin(c)[cos() ] + sin() cos(c) sin(c)[cos() ] sin() cos(c) + lim 0 ( ) 0 ( cos() = sin(c) lim + cos(c) lim 0 = sin(c) 0 + cos(c) = cos(c). ) sin(). (a) Use inuction to prove tat x (xn ) = nx n for all n N. Hint: Write x n+ = x n x an use te prouct rule. Solution. Wen n =, we ave (x) = = x x0, so te statement is true. Now suppose te statement ols for some n. Ten, by te prouct rule, x (xn+ ) = x (xn x) = nx n x + x n = nx n + x n = (n + )x n. (b) Use eiter te quotient rule or cain rule, togeter wit te result of part (a), to prove tat te erivative of x (xn ) = nx n for all negative integers n. Solution. If n Z an n < 0, ten m = n N, so by (a) x xm = mx m. Tus, by te quotient rule x (xn ) = x x = mxm = mx m = nx n. m (x m ) 2 5. Let f(x) = x sin(/x) for x 0 an f(0) Sow tat f is ifferentiable everywere except x 2

3 Solution. For x 0, te prouct rule an cain rule imply tat f is ifferentiable wit erivative f (x) = x cos(/x) ( /x 2 ) + sin(/x). On te oter an f (0) f(x) f(0) x 0 x sin(/x) x sin(/x). As sown in class, tis limit oes not exist, so f is not ifferentiable at x 6. Let f(x) = x 2 sin(/x) for x 0 an f(0) (a) Sow f is ifferentiable everywere (incluing x = 0). Solution. For x 0, te prouct rule an cain rule imply tat f is ifferentiable wit erivative Next, f (x) = x 2 cos(/x) ( /x 2 ) + 2x sin(/x) = 2x sin(/x) cos(/x). f (0) f(x) f(0) x 0 x 2 sin(/x) x x sin(/x). Since x sin(/x) x for all x 0 an lim x x = 0, te squeeze teorem implies tat lim x sin(/x) = 0, so f is ifferentiable at x = 0 wit f (0) (b) Sow f is not continuous at x Solution. Since f (0) = 0, in orer for f to be continuous at x = 0 we woul nee lim f (x) As above, by te squeeze teorem, we know lim 2x sin(/x) Tus, since lim cos(/x) oes not exist, it follows tat lim f (x) oes not exist. Hence f is not continuous at x 7. (a) Prove tat te polynomial p(x) = x 3 3x 2 + 6x + 2 as exactly one real root. Solution. Since p is a polynomial, p is continuous on te interval [ 2, 0]. Tus, since p( 2) = 20 < 0 an p(0) = 2 > 0, te Intermeiate Value Teorem implies tere exists c [0, 2] suc tat p(c) Tis proves p as a real root. To see tat p as only one real real, suppose tere exists a secon root c 2. Ten since p(c) = p(c 2 ) = 0, Rolle s Teorem implies tere exists some point c 3 between c an c 2 suc tat p (c 3 ) But p(x) = 3x 2 6x+6 = 3(x ) 2 +3, so p(x) 0 for all x an tus p cannot ave two roots. (b) Sow tat te equation x 2 = 3 + sin(x) as exactly two solutions. Solution. Te equation is equivalent to f(x) = 0 were f(x) = x 2 sin(x) 3. We first sow tat tis equation can ave at most two solutions. Suppose instea tat tere exist c < c 2 < c 3 suc tat f(c ) = f(c 2 ) = f(c 3 ) Since f 3

4 is continuous on [c, c 2 ] an ifferentiable on (c, c 2 ), te Mean Value Teorem implies tat tere exists c (c, c 2 ) suc tat f (c ) = f(c 2) f(c ) c 2 c = 0 c 2 c Similarly, tere exists c 5 (c 2, c 3 ) suc tat f (c 5 ) Next, since f is continuous on [c, c 5 ] an ifferentiable on (c, c 5 ), te Mean Value Teorem implies tat tere exists c 6 (c, c 5 ) suc tat f (c 6 ) = f (c 5 ) f (c ) c 5 c = 0 c 5 c But tis is impossible since f (x) = 2 + sin(x) for all x. Next we sow tat f oes in fact ave two roots. Since f(0) = 3 < 0 an f(3) = 6 sin(3) > 5 > 0 an since f is continuous on [0, 3], te Intermeiate Value Teorem implies tat tere exists c (0, 3) suc tat f(c ) Since f( 3) = 6 sin( 3) > 5 > 0 an f is continuous on [ 3, 0], te Intermeiate Value Teorem implies tat tere exists c 2 ( 3, 0) suc tat f(c 2 ) Since c > 0 an c 2 < 0, tese roots are istinct. (c) Prove tat te equation x = 3x 3 + as exactly two solutions. Solution. Rewrite te equation as x 3x 3 = 0 an set f(x) = x 3x 3. Ten x is a solution of te equation if an only if f(x) Since f is a polynomial, f is continuous everywere. Since f(0) = < 0 an f() = 63 > 0, te Intermeiate Value Teorem implies tere exists c [0, ] suc tat f(c ) Since f( ) = 3 > 0 an f(0) = < 0, te Intermeiate Value Teorem implies tere exists c 2 [, 0] suc tat f(c 2 ) To see tat tese are te only two solutions, first notice tat f (x) = x 3 9x 2 = x 2 (x 9). Tus f (x) > 0 for all x > 9, an terefore f is strictly increasing on ( 9, ). Tis implies f can ave at most one root in te interval ( 9, ), namely c 2. Next, f (x) < 0 for x (, 0) or x (0, 9 ). Tus f is strictly ecreasing on eac of tese intervals, an can ave at most one root in eac interval. But f(0) =, so f(x) < 0 for all x (0, 9 ) an tus f as no roots in tat interval. Te point c must be te one root of f in te interval (, 0). Hence f as exactly two roots. 8. Suppose f is ifferentiable everywere an f (x) 2 for all x R. Sow tat tere is at most one number c suc tat f(c) = 2c + 3. Solution. Let g(x) = f(x) 2x. Ten g is ifferentiable everywere an g (x) = f (x) 2 0 for all x R. Suppose tere exist real numbers c < c 2 suc tat f(c ) = 2c + 3 an f(c 2 ) = 2c Ten g(c ) = g(c 2 ) = 3. By te Mean Value Teorem, tere exists c (c, c 2 ) suc tat g (c) = g(c 2) g(c ) c 2 c Tis contraicts te fact tat g (x) 0 for all x.

5 9. Suppose f is twice ifferentiable on R, f(0) = 5, f(2) = 8 an f(7) = 2. (a) State two values tat f (x) must attain for x (0, 7). Solution. Applying te Mean Value Teorem on te interval [0, 7] implies tere exists c (0, 7) suc tat f (c ) = f(7) f(0) 7 0 = 3 7. Applying te Mean Value Teorem on te interval [0, 2] implies tere exists c 2 (0, 2) suc tat f f(2) f(0) (c 2 ) = = Applying te Mean Value Teorem on te interval [2, 7] implies tere exists c 3 (2, 7) suc tat f f(7) f(2) (c 3 ) = = Also, since f is continuous on [0, 7], te Extreme Value Teorem implies tat f attains a max an a min on [0, 7]. Since f(2) = 8 is greater tan f(0) an f(7), te maximum value is attaine at some point c (0, 7). By Teorem.3.3, f (c ) (b) Sow tat tere is at least one point c (0, 7) wit f (c) < 0. Solution. Since c 2 < c 3, applying te Mean Value Teorem to f on [c 2, c 3 ] implies tat tere exists some c (c 2, c 3 ) suc tat f (c) = f (c 3 ) f (c 2 ) c 3 c 2 = c 3 c 2 < 0. 5