Math 1000 Final Exam Review Solutions. (x + 3)(x 2) = lim. = lim x 2 = 3 2 = 5. (x + 1) 1 x( x ) = lim. = lim. f f(1 + h) f(1) (1) = lim

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1 Math Final Eam Review Solutions { + 3 if < Consider f() Find the following limits: (a) lim f() + + (b) lim f() (c) lim f() does not eist Find each of the following limits: + 6 (a) lim (b) lim + ( + 3)( ) ( + + ) ( + + ) ( + ) ( + + ) ( + + ) + + (c) sin() lim tan(3) sin() sin(3) cos(3) sin() 3 sin(3) cos(3) 3 sin() sin(3) cos(3)3 3 sin() 3 sin(3) cos(3) (a) Give the limit definition of the derivative The derivative of a function f at a number a is f f(a + h) f(a) (a) h h (b) Use the limit definition of derivative to find f () if f() + 3 f f( + h) f() () h h + h h h + h ( + h + ) h h ( + h + ) ( + h) h h( + h + ) h h h( + h + ) h + h +

2 (a) State the Mean Value Theorem Let f be a function that satisfies the following conditions: (i) f is continuous on the closed interval [a, b] and (ii) f is differentiable on the open interval (a, b) Then there is a number c in (a, b) such that (b) State Rolle s Theorem f (c) f(b) f(a) b a Let f be a function that satisfies the following conditions: (i) f is continuous on the closed interval [a, b], (ii) f is differentiable on the open interval (a, b), and (iii) f(a) f(b) Then there is a number c in (a, b) such that f (c) (c) Let f() on the interval [, ] Show that the Mean Value Theorem applies and then find all the values of c such that f (c) f(b) f(a) b a Since f is a polynomial, it is continuous on [, ] and it is differentiable on (, ) Thus, the Mean Value Theorem applies Differentiating f () By the Mean Value Theorem, we know that f (c) c 5 Differentiate the following functions: f() f( ) ( ) () ( ) ( ) 3 c (a) H() e + cos(a) + arcsin(), where a is a constant (b) f(t) t t + H () e + e + cos(a) a sin(a) + f (t + )(t) (t )(t) (t + ) t3 + t t 3 + t (t + ) t (t + )

3 (c) g() arctan ( + e ) g + ( + e ) ( + e ) / ( + e ) + e + e ( + + e ) (d) f() ln(sin( )) f () sin( ) cos( )() cos( ) sin( ) 6 (a) Differentiate implicitly to find the derivative of y if + y 3 / + y / y y / y / y / y / y (b) Find the equation of the tangent line of the function in part (a) at the point (, ) The slope of the tangent line at the point (, ) is y Then the equation of the tangent line is y ( ) y Differentiate with respect to (use logarithmic differentiation): (a) y e ( ) 3/ + First, take the natural log of both sides ( ) e ( ) 3/ ln(y) ln + ln(e ( ) 3/ ) ln( + ) ln() + ln(e ) + 3 ln( ) ln( + ) ln() ln( ) ln( + ) 3

4 Differentiating implicitly, we obtain (b) y ( + ) / y y ( y y ) + ( e ( ) 3/ ) + Take the natural log of both sides Differentiating implicitly, we obtain ln(y) ln( + ) y y ( ) ln( + ) + + ( ) ln( + ) y y + ( + ) ( ) ln( + ) ( + ) / + ( + ) 8 A ladder ( feet long) is leaning against the wall of a house The bottom of the ladder is sliding away from the wall at a rate of feet/sec How was is the top of the ladder moving down the wall, when the bottom of the ladder is feet away from the wall? y 5 Let be the distance that the bottom of the ladder is away from the wall Let y be the distance the top of the ladder is from the ground So we are given that d ft/sec and we want to find dy when Since the ladder makes a right-angled triangle with the wall, we have + y 5

5 When, y 5 (5 )(5 + ) 9 7 Differentiating our equation with respect to t, we obtain Solving for dy, So, when, y 7 and d y dy dy + y dy d d y dy () () Find the linearization L() of f() + 3 for a Recall the formula for the linearization of f: L() f(a) + f (a)( a) f(a) f( ) ( ) + 3( ) + 3 f () f (a) f ( ) ( ) 3 + 6( ) 6 Then L() + ( )( ( )) L() 6 Find the following limits: (a) e lim H e H e form form (b) lim / Let y / Then, taking the natural log, ln(y) ln( / ) ln() 5

6 Then this limit has the form We use L Hospital s rule: ln() lim H Therefore, y / e Given the function f() + 3 3, answer the following: (a) What is f ()? f 3 3 (b) Find the critical numbers Set f to zero and solve for : 3 3 ± The critical numbers are and (c) Find the intervals of increasing/decreasing < < < > f + f < on (, ) and (, ) so f is decreasing on (, ) and (, ) f > on (, ) so f is increasing on (, ) (d) Find the local maimum and/or the local minimum value f changes from increasing to decreasing at, so f has a maimum value of f() f changes from decreasing to increasing at, so f has a minimum value of f( ) (e) Find the intervals of concavity First, we find the second derivative: f 6 Setting f to zero and solving for, we get f > for <, so f is concave up on (, ) f < for >, so f is concave down on (, ) (f) What is the inflection point? Since f changes concavity at, (, f()) (, ) is an inflection point 6

7 Given the function f(), answer the following: (a) What is the domain? When, the function is undefined So our domain is all in the real numbers such that We can write this as D (, ) (, ) (b) is not in the domain, so there are no y-intercepts Setting y, we get that is an intercept Therefore, (, ) is an -intercept (c) Are there any vertical asymptotes? Are there any horizontal asymptotes? If so, what are they? lim, so is a vertical asymptote lim / / Thus, y is a horizontal asymptote lim / / This just gives the same horizontal asymptote (d) Find the critical numbers Taking the derivative: Set f : f ()( ) ( )() ( ) ( ) 3 ( ) + Therefore, is a critical number is not a critical number because it is not in the domain (e) Find the intervals of increasing/decreasing < < < > f + 7

8 f < on (, ) and (, ), so f is decreasing on (, ) and (, ) f > on (, ), so f is increasing on (, ) (f) What are the local maimum value and the local minimum value? Since f changes from increasing to decreasing at, f has a maimum value of f() f does not have any maima or minima at since f has a vertical asymptote at (g) Find the intervals of concavity First, we find the second derivative: f ( )(3 ) ( )(3 ) ( 3 ) 3 (6 3 3 ) ( 3) 6 6 ( 3) Setting f to zero and solving for, we get 3 < < < 3 > 3 ( 3) f + Since f < on (, ) and (, 3), f is concave down on (, ) and (, 3) Since f > on (3, ), f is concave up on (3, ) (h) Find the inflection points Since f changes concavity at 3, (3, f(3)) (3, ) is an inflection point 9 (i) Using parts (a)-(g), graph the function 8

9 3 Find two numbers whose difference is and whose product is a minimum Let our two numbers be and y We are given that y and we want to minimize their product: P y First, y + Then we rewrite P ( + ) + Differentiating P + Setting to zero and solving for, we obtain P + 5 Taking the second derivative, we have P > Since P ( 5) and P >, by the Second Derivative Test, P has a minimum at 5 Then, y + + ( 5) 5 So our two numbers are 5 and 5 (a) Estimate the area under the curve of f() on the interval [, ] using approimating rectangles and right endpoints R f( ) + f() + f(3 ) + f() ( ( ) + () + ( 3 ) ) + () ( ) 3 7 (b) State the definition of the definite integral of f from a to b (as the limit of a Riemann sum) The definite integral of f from a to b is where b a n b a f() d n and i are the sample points n f( i ) (c) Write ( ) d as the limit of the Riemann sum over n intervals (Do not evaluate the limit) First, We can take the sample points to be the right endpoints (it n n makes no difference to the infinite limit): i a + i + i i n Then, using the definition from part (b), ( ) d n 9 i ( n i ( ) ) i ( n n ) n

10 5 Evaluate the following integrals (a) 9 d 9 9 d / / d [ ] 3/ 9 [ ] 9 3/ / / 3 3/ / ( ) ( ) 3 93/ (9 / ) 3 ()3/ () / (b) (c) (d) π/ e sin(e ) d cos() d [sin()] π/ sin( π ) sin() sec () d tan() + C Use a substitution: let u e, then du e d Then e sin(e ) d sin(e ) e d sin(u)du (e) d cos(u) + C cos(e ) + C Use a substitution: let u + 3, then du d Also, changing the limits of 3 integration: u and 7 u 5 Then d u du 3 [ ] / 9 (53/ 3/ ) 9 (53 3 ) 9 (7) 6