Continuity and Differentiability Workseet (Be sure tat you can also do te grapical eercises from te tet- Tese were not included below! Typical problems are like problems -3, p. 6; -3, p. 7; 33-34, p. 7; -4, p. 3; 4, 46-48, 5 p. 76). Definition: A function f is said to be continuous at a if: f(a). Te definition of continuity implies tat we ave tree tings to ceck. Wat are tey? () f(a) eists (or f(a) is defined), () f() eists. (3) Te numbers in () and () are te same. 3. Finis te definition: A function f is said to be continuous on te interval [a, b] if: f is continuous for every point in (a, b), is left continuous at a and rigt continuous at b. 4. Finis te definition: Te derivative of f at a is: f (a) f(a + ) f(a) or f() f(a) a 5. Finis te definition: A function f is said to be differentiable on te interval (a, b) if: f is differentiable at eac in te interval (a, b). 6. Wy is te interval open in te last definition? Because we need to be able to take te it (from bot sides) of eac number in (a, b). If we included te point a or b, we d ave to take a one-sided derivative. 7. List tree interpretations of te derivative of f at a. () Slope of te Tangent Line to f at a. () Velocity (if f is a displacement function over time), (3) Instantaneous rate of cange of f. 8. True or False, and give a sort reason: (a) If a function is differentiable, ten it is continuous. Tis is true (it was a teorem in class). (b) If a function is continuous, ten it is differentiable. False. Eample: y at 0. (c) If f is continuous on [, ] and f( ) 4 and f() 3, ten tere is an r so tat f(r) π. True- Tis is te statement of te Intermediate Value Teorem. (d) If f is continuous at 5, and f(5), ten te it as of f(4 ) must be. True. Because f is continuous, f(a). (e) All functions are continuous on teir domains. Not True- All of our basic functions are, but tere are functions tat are not continuous anywere (for eample, te function tat is zero on te rationals and on te irrationals). 9. Were is eac function continuous? 4 (a) Tis function will be continuous on its domain because it is constructed from te functions 4, and. To find te domain, we require tat: so we use a table to solve: ( + )( ) ( + )( ) 0 ( + ) + + + + ( ) + + + + ( + ) + + + ( ) + + + < < < < < < < > By te table, te domain is:, or < <, or. wic is also were f() is continuous.
(b) sin ( ) First, recall tat te domain of sin () is. Terefore, te domain of sin ( ) is were. Tis implies tat 0, or were 0. Tus, te answer is: Te function f is continuous on its domain,. ( ) + 3 (c) ln 5 Tis function will be continuous on its domain- Te function ln() as a domain: > 0, so ln( +3 5 ) +3 as a domain tat satisfies: 5 > 0. Use a table to solve: + 3 + + 5 + < 3 3 < < 5 > 5 Conclusion: Te function is continuous for < 3 or > 5. (d) + 5 + 6 Here, we need to make sure tat + 5 + 6 0, so solve by factoring. Conclusion: Te function is continuous on all reals ecept were 3 and. 0. Eplain wy te function is discontinuous at te given point, a. (a) ln + 3 at a 3 (Etra: Is f continuous everywere else?) f is not continuous at a 3 because f is not defined for a 3. (Yes, f is continuous for all oter ). (b) { 8 4, if 4 3, if 4 a 4 For tis function, () f is defined at a 4, and f(4) 3. () + 6. (3) Te 4 4 answers for () and () are not te same, so f is not continuous at a 4. (c) +, at a For tis function, f is not continuous at a because f( ) is not defined. (d) {, if, if > a We again ceck te tree properties: () f(), so f is defined at a. () f() does not eist- Te it coming from te rigt is 0, te it coming from te left is. We don t ave to ceck te tird property- f is not continuous because te it at a does not eist.. For eac function, determine te value of te constant so tat f is continuous everywere: (a) { 6 4, if 4 C, if 4 First, f(4) C, so tat does not restrict our coice of C. Net, we want te it to eist (and be 6 equal to C), so: C 4 4 8.
(b) 3, if < 0 c + d, if 0 + 8, if > Again, we want to make sure te its matc at 0 and at as we come in from te rigt and left. At 0: So d. Net, we ceck : d and 0 + 0 (c) so c. 3 and + c + { 7+ 6+4, if 7, and k, if We want te its from te left and rigt of to matc up. First from te left: 7 + 6 + 4 7 + 6 + 4 7 + + 6 + 4 7 + + 6 + 4 and, if we take te it from te rigt: 7 + + 6 + 4 8 k k + Putting te two togeter, k 8. If f and g are continuous functions wit f(3) 4 and 3 [f() g()] 5, wat is g(3)? By continuity, [f() g()] f() g() f(3) g(3) 8 g(3) 3 3 3 so tat we now ave: so g(3) 3. 8 g(3) 5 3. Sow tat tere must be at least one real solution to 5 4 0. Tis is an Intermediate Value Teorem, were 0 is te Intermediate Value. Terefore, we need to find an so tat 5 4 < 0 and an so tat 5 4 > 0. For eample, if, ten we get a 4. If, we get 3 4 4 4 > 0. Terefore, a solution to te equation is somewere between and. 4. Eac it is te derivative of some function at some number a. State f and a in eac case: + (a) at a. (b) 9 9 at a 3
( sin π (c) + t) t 0 t sin() at a π 5. For eac function below, compute te derivative using te definition. Also state te domain of te original function, and te domain of te derivative function. (a) + Domain of f: + ( + ) + + + ( + ( + ) + + ) + ( + ) + + ( + ) + + + ( + ) + + ( + ( + ) + + ) + (b) g() (+) (+) (+) ( + ) 3 (c) () + [ + + + ] [ + ( ) ] + + + + + + + + + + + ( + + ) + (d) 3 (e) 6. Let. + (+) 3 (+) 3 ( 3 3 ) 3 3 3 3 ( 3 + 3 ) f() f(a) (a) Use a 3 + 3 3 + 3 (3 ) 3 ) (3 ) 3/ ( + )( ) (( + ) ) + (( + ) )( ) (( + ) )( ) ( ) to compute f (a), for a 0. a a a ( a) ( + a) + a ALTERNATIVE: You could also multiply numerator and denominator by + a and get te same result. (b) Sow tat f (0) does not eist. Wat does tis mean wit respect to te grap of f at a 0? From our formula, we see tat, as 0, f (), wic means tat at 0, tere is a vertical tangent line. 4
7. Given f below, were is f not continuous? 0, if 0 5, if 0 < < 4 5, if 4 Not continuous at: 0, because te it from te rigt is not te it from te left. It is continuous at 4, and for all oter. 8. Let 3. (a) Find f (). (b) Compute te equation of te line tangent to f at te point (, 4). f( + ) f() ( + ) 3 ( + ) ( 3 ()) (0 + 6 + ) 0 9. Sketc te grap of a function tat satisfies te following conditions: g(0) 0, g (0) 3, g () 0, g () Your grap sould ave at least: A point at (0, 0) wit te curve going troug te origin fairly steeply (local slope of 3), were te curve goes troug, te curve sould be flat (slope of zero), and finally, were te curve goes troug, te slope sould be about. 0. Find te slope of te line tangent to y + at 3, ten compute te equation of te line. f( 3 + ) f( 3) ( 3 + ) + ( 3 + ) 3 4 + m tan 4 Te tangent line as te equation: (In general: y f(a) f (a)( a)) y 3 4( + 3) 5