(a) At what number x = a does f have a removable discontinuity? What value f(a) should be assigned to f at x = a in order to make f continuous at a?

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1 Solutions to Test 1 Fall 016 1pt 1. Te grap of a function f(x) is sown at rigt below. Part I. State te value of eac limit. If a limit is infinite, state weter it is or. If a limit does not exist (but is not infinite), explain wy not. (a) lim x 1 f(x)= (b) lim x f(x)= (c) lim x 4 f(x)= 1 (d) lim x 4 +f(x)= (e) lim x 4 f(x) d.n.e. because lim x 4 f(x) lim x 4 + f(x) Part II. Use te Limit Laws to find lim x 4 +(x+1)f(x). Sow your work. lim (x+1)f(x) (x+1) lim f(x) = (4 +1)( ) = 10 x 4 + x 4 + x 4 + Part III. (a) At wat number x = a does f ave a removable discontinuity? Wat value f(a) sould be assigned to f at x = a in order to make f continuous at a? a = 1 f(a) = (b) At wat number x = b does f ave a jump discontinuity? b = 4 (c) At wat number c is f continuous but not differentiable? c = 1

2 13pt. Find te following limits. If a limit is infinite, determine weter it is or. Sow your work. Please note tat you MAY NOT use L Hospital s Rule ere (even if you know wat it is and ow to use it). (a) lim x arctan(x4 ) Let t = x 4. We ave lim t =. Ten lim x x arctan(x4 ) arctan(t) = π t (b) lim x 3 x+1 x 3 x + 1 We ave lim (x 3) = 0, but lim (x + 1) = 4 0. Terefore, lim is infinite. x 3 x 3 x 3 x 3 Te numerator is positive, but te denominator is negative (x < 3), terefore, x + 1 lim x 3 x 3 =. (c) lim x 1 x 5 x 1 x 5 x 1 + x 1 + x 1 4 (x 5)( x 1 + ) x 5 (x 5)( x 1 +) 1 x 1 + = = 1 4 (d) lim x e x + 4e x 1 Note tat lim x e x = 0. Ten lim x e x + 4e x 1 = =

3 6pt 3. (A) Find te limit lim (3+) (6 +) (6 +) = 6 (B) Te limit in part (A) represents te derivative of some function f at some number a. Find suc an f and a f(x) = x a = 3 (Oter solutions are possible.) 7pt 4. Usetegridbelowtosketc tegrapofafunctionf(x)tatsatisfiesalloftegivenconditions. f is continuous on (, ) except at 3 and lim x 3 f(x) =, lim x 3 +f(x) = lim x f(x) =, lim x +f(x) = 4 f(x) is continuous from te rigt at x = lim x f(x) =

4 8pt 5. Let f(x) = ( ) πx sin if 0 x < 1 x+1 a if x = 1 x+8 b if x > 1 were a and b are constants. (a) Find lim x 1 f(x) and lim x 1 +f(x) (please note: your answer may depend on b). ( ) πx lim f(x) sin = sin(π/) = 1 x 1 x 1 x +1 lim f(x) ( x + 8 b) = 1 +8 b = 3 b x 1 + x 1 + (b) For wat value of b does te limit lim x 1 f(x) exist? lim f(x) exists if (and only if) lim f(x) f(x). i.e., if 1 = 3 b, tat is if b =. x 1 x 1 x 1 + (c) Find te values of a and b (if any) for wic f(x) is continuous at x = 1. f(x) is continuous at 1 if lim x 1 f(x) exits and is equal to f(1). Ten b = (for te limit to exist) and a = f(1) x 1 f(x) = 1. Tus, a = 1, b =. pt 6. Te grap of a function f(x) defined on [0,3] is sown below. Tere is no number c in (0,3) suc tat f(c) =. Explain clearly wy te Intermediate Value Teorem may not be applied to tis function f on [0,3]. Te Intermediate Value Teorem may not be applied to f on te interval [0,3] because f is not continuous on te interval [0,3] (f as a discontinuity at 1).

5 13pt 7. Let f(x) = x x 6 (x+)(x 1). (a) Find lim f(x) and lim f(x). Sow your work. x x 1 1 lim f(x) 6 x x (ere we divided te numerator x x (1 + )( 1 ) and denominator by x ) x x = (1 +0)( 0) = 1 and similarly, 1 1 lim f(x) 6 x x x x (1 + )( 1 ) = 1 x x (b) Find te limit if it exists. Sow your work. x x 6 (i) lim f(x) x x (x + )(x 1) (x+)(x 3) x (x+)(x 1) x x 3 x 1 x 5 5 = 1 (ii) lim x 1 f(x) x 1 x x 6. Te denominator approaces 0 wile (x +)(x 1) te numerator approaces Terefore, te limit is infinite. Since bot, te numerator and te denominator, are negative, te limit is positive. (c) Find equations of te orizontal and vertical asymptotes of te grap of f(x) HA: y = 1 VA: x = 1 (since lim f(x) is infinite) x 1 pt 8. Yes or No? Circle your answers. (a) Are te functions x 3x+ x 1 and x te same? YES NO (b) Are te limits of x 3x+ x 1 and x as x 1 te same? YES NO

6 6pt 9. Find te derivative f (x) of te function f(x) = 1 using te limit definition of derivative. x+1 Please note: you may NOT use differentiation rules ere. f (x) f(x+) f(x) 1 (x+)+1 1 x+1 x+1 (x++1) (x++1)(x+1) x + 1 x 1 (x+ +1)(x +1) (x+ + 1)(x+1) (x+ + 1)(x+1) = (x + 1) = f (x) = (x+1) IN PROBLEMS 10 1 YOU MAY USE DIFFERENTIATION RULES. 14pt10. Differentiate. Sow your work. (a) f(t) = t 1/3 + 3 t + 41/3 f(t) = t 1/3 +3t + 4 1/3 = f (t) = 1 (b) g(x) = x+3. Simplify your answer. x+1 3 t /3 + 3 ( )t = 3 t /3 6t 3 g (x) = (x+1) 1 (x + 3) ( + 0) (x + 1) = x + 1 x 6 (x + 1) = 5 (x +1) (c) (θ) = θ tanθ (θ) = θ sec θ + tanθ 1

7 IN PROBLEMS 10 1 YOU MAY USE DIFFERENTIATION RULES. 10pt11. Let f(x) = cosx e x. (a) Find f (x). f (x) = ex ( sinx) cosxe x e x = ex ( sinx cosx) e x = sinx cosx e x (b) Find an equation of te tangent line to te curve y = f(x) at te point were x = 0. Give your answer in te slope-intercept form. Te point-slope form of te equation: y f(0) = f (0)(x 0) We ave f(0) = cos0 e 0 = 1 1 = 1 and f (0) = sin0 cos0 = 0 1 e 0 1 = 1. Ten te equation is y 1 = 1(x 0) or, in te slope-intercept form, y = x + 1. (c) Find te number a in te interval ( π/, π/) at wic te tangent line to te grap of f is orizontal. m tan = 0 = f (a) = 0 = sina cosa e a = 0 = sina cosa = 0 = sina = cosa = tana = 1 (we divided bot sides by cosa since cosa 0 for a satisfying te equation) = a = π 4. (Tere are infinitely many solutions, but only tis one is in te interval ( π/,π/).)

8 IN PROBLEMS 10 1 YOU MAY USE DIFFERENTIATION RULES. 1pt1. Te displacement (in feet) of a particle moving in a straigt line is given by te equation of motion s(t) = t +1, were t is measured in seconds. t Answer te following questions. (a) Find te average velocity of te particle over te time interval 1 t 3. Please include units of measurement in your answer. v aver (1,3) = s(3) s(1) 3 1 = = 10 3 = 4 6 = 3 ft/s. (b) Find te instantaneous velocity v and acceleration a of te particle at time t (int: simplify s(t) first). s(t) = t + 1 t = v(t) = s (t) = 1 1 t = 1 1 t and a(t) = v (t) = ( )t 3 = t 3 (c) Find v(1) and a(1). Please include units of measurement in your answers. v(1) = = 0 ft/s and a(1) = 1 3 = ft/s