Solution: As x approaches 3, (x 2) approaches 1, so ln(x 2) approaches ln(1) = 0. Therefore we have a limit of the form 0/0 and can apply the.

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1 MATH, solutions to practice problems for the final eam. Compute the it: a) 3 e / Answer: e /3. b) 3 ln( 3) Answer:. c) 3 ln( ) 3 Solution: As approaches 3, ( ) approaches, so ln( ) approaches ln() =. Therefore we have a it of the form / and can apply the L Hôpital s rule: d) (3 )( 3 3) ln( ) 3 3 = 3 /( ) =. Solution: Let us divide the top and the bottom of this fraction by 5 : (3 )( 3 3) = 8 7/ + 9/ 5 (3 / )( 3/ 3 ) = 8 + (3 )( ) = 4 3. e e) 3 Solution: As this is the it of the type /, we can apply the L Hôpital s rule several times: e = e 3 3 = e 6 = e 6 =. f) arctan() 3 Solution: As this is the it of the type /, we can apply the L Hôpital s rule several times: arctan() 3 = /( + ) 3 3 ( + ) = = 3 ( + ) = 3( + ) = 3.

2 . Compute the derivative of the following functions: a) f() = ln Answer: ln() b) f() = e 3 Answer: 6e 3. c) f() = ( ) 5 cos Answer: 5( ) 4 cos ( ) 5 sin. d) f() = sin(ln ) Answer: cos(ln ). e) f() = e3+ cos + Answer: (3 cos +6+sin )e3+ (cos +). 3. Find the minimal and maimal values of a function: a) f() = e on [, ] Solution: We have f () = e e = ( )e, so f () = for = and =. Since = is not on the interval and = is one of the endpoints, we just need to compute f() = with f() = e. Therefore the minimal value is and the maimal value is e. b) f() = on [, ] Solution: We have f () = 6 6 = 6( ), so f () = for = and =. We need to compute f( ) = 3+ = 4, f() =, f() = 3+ =, f() = 6 + = 5. Therefore the minimal value is ( 4) and the maimal value is 5. c) f() = sin on [, π] Solution: We have f () = sin cos, so f () = when either sin = (so = πk) or cos = (so = π/ + πk). On the interval [, π] we have 3 critical numbers, π/, π and both endpoints are among them. Therefore one needs to compute f() =, f(π/) =, f(π) =, and the minimal and maimal values are and respectively. d) + on [, ]

3 Solution: We have f () = + () = ( + ), and the critical (+ ) numbers are = and =. We have f( ) = /5, f( ) = /, f() = /, f() = /5. Since / > /5, the minimal and maimal values are / and / respectively. 4. Find the equation of the tangent line to the graph of f() = ln at = 5. Answer: y = + (ln 5 ) For a given function: Find the domain Determine the equations of vertical and horizontal asymptotes Find the derivative and determine the intervals where the function is increasing/decreasing Find the second derivative and determine the intervals where the function is concave up/down, find inflection points Draw the graph using all the information above a) f() = Solution: The function is defined everywhere, and there are no asymptotes. We have f () = 6 6 = 6( ), so the function is increasing for > and <, and decreasing for < <. Furthermore, f () = 6, so the function is concave up for > / and concave down for < /, and it has an inflection point at = /. 3

4 3 b) f() = e Solution: The function is defined everywhere. To find the asymptotes, remark that at the function e goes to infinity, so e =. To find the it at +, let us use the L Hôpital s rule: + e = + e = + e =. Therefore the graph has a horizontal asymptote y = at +. Now f () = e e = ( )e, so the function increases for < and decreases for >. Furthermore, f () = e ( )e = ( )e, so the function has an inflection point at = c) f() = ln( + ) 4

5 Solution: Since +, the function is defined and nonnegative everywhere. As approaches ±, + +, so ln( + ) +, and there are no horizontal asymptotes. Now f () =, so the function decreases for < and increases for + >. Furthermore, f () = ( + ) () ( + ) = ( ) ( + ), and the function has inflection points at = ± d) f() = + Solution: The function is defined for, and has a vertical asymptote at =. To find horizontal asymptotes, we have + = / + / =. Now f () = + ( ) (+) = (+), so the function increases on every interval where it is defined. Furtermore, f () = 4( + ) 3, so the function is concave down for > and concave up for <. 5

6 Find the integral: a) e d Solution: Let u =, then du = d, d = /du, and e d = e u du = eu + C = e + C. b) ( 3) d Solution: Let u = 3, then du = d and ( 3) d = u du = u3 ( 3)3 + C = C. c) ( 3) d Solution: Let us epand ( 3) : ( 3) d = ( 6 + 9) d = ( 5/ 6 3/ + 9 / )d = d) sin( + 3)d 7 7/ 5 5/ / + C. Solution: Let u = + 3, the du = d, and sin( + 3)d = sin(u)du = cos(u) + C = cos( + 3) + C. 6

7 e) d ln Solution: Let u = ln, then du = d and ln d = du = ln(u) + C = ln(ln ) + C. u f) + d + Solution: We have d = + d = + ( + )d = + arctan + C Find the definite integral: a) 3 d + Solution: We have 3 b) 3 + d = arctan() 3 = arctan(3) arctan() = arctan(3). d + Solution: Let u = +, then du = d, so 3 + d = 5 u du = ln(5) (ln(5) ln()) =. c) π /4 cos( ) d Solution: Let u =, then du = d, so d = du and π /4 d) 5 eln d Solution: cos( ) d = 5 π/ e ln d = 5 cos(u)du = (sin(π/) sin()) =. d = 5 = 5 7 =.

8 e) 5 (ln ) 5 d Solution: Let u = ln, then du = d and 5 8. Find the area (ln ) 5 d = ln 5 u 5 du = u6 6 ln 5 = (ln 5)6. 6 a) Between the parabola y = and the line y = 3 Solution: Let us find the intersection points: = = ( )( ) =, so the intersection points are = and =. Now A = (3 )d = ( ) = 6 4 8/3 3/++/3 = /6. b) Between the lines =, y = and y = 3. Solution: Let us find the intersection point between the second and the third line: = 3 3 = 3 =. Therefore A = (3 )d = (3 3) = (3 3 /) = 3 3/ = 3/. c) Between the hyperbola y = / and the line y = 5 Solution: Let us find the intersection points: = 5 = 5 5 +,, = 5/ ± 5/4 4 so = /, =. = 5/ ± 3/, 8

9 Now A = / ( 5 )d = (5 / ln()) / = 5 ln() 5/4 + /8 + ln(/) = 5/8 ln(). d) Between the parabolas y = 3 and y = 5 3. Solution: Let us find the intersection points: 3 = = 8 =, so the parabolas intersect at = ±. We have A = ( )d = (8 4 ) = (8 4 3 /3) = = 8 8 /3 8( ) + 8( )/3 = 3 /3. 9. Find the volume of the body obtained by the rotation around -ais of the following region bounded by the lines =, = 4 and the graph of: a) y = Solution: The sections of this body are circles with radius, the area of such a circle equals π( ). We have: V = π( ) d = π d = π / 4 = π 6 = 5π. b) y = Solution: Similarly, V = π( ) d = π 4 d = π 5 /5 4 = π 45 5 = 3π. 5 c) y = + 9

10 Solution: Similarly, V = π( + ) d = π ( + ) d = π( + ) 3 /3 4 = π = 7π 3. d) y = e + Solution: Similarly, V = π(e + ) d = π (e + e + )d = π(/e + e + ) 4 = = π(e 8 / + e /e e ).. Consider the function { +, if < f() = + a + b, if. a) For which values of the parameters it is continuous? Solution: The function is continuous, if its its from the left and from the right at = coincide: ( ) + = ( ) + a( ) + b = a + b b = a. b) For which values of the parameters it has a derivative at every point? Solution: The function clearly has a derivative for all, and it has derivative at = if it is continuous at this point (see (a)), and the derivatives from the left and from the right coincide: = ( ) + a, so a = 3 and b =. 3 3 c) For all a and b, find the integral 3 f()d. 3 Solution: f()d = 3 (+)d+ 3 ( +a+b)d = ( /+) 3+( 3 /3+a /+b) 3 = / 9/ /a + 3b + /3 /a + b = /3 + 4a + 4b.