Math 1210 Midterm 1 January 31st, 2014

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1 Mat 110 Midterm 1 January 1st, 01 Tis exam consists of sections, A and B. Section A is conceptual, wereas section B is more computational. Te value of every question is indicated at te beginning of it. You may only use scratc paper and a small note card. No cell pones, calculators, notes, books or music players are allowed during te exam. Name: UID: Section A: Conceptual questions. 1. (i) ( points) Wat does it mean for a function f(x) to be continuous at x = c? A function is continuous at x = c if x x f(x) = f(c), namely: (i) f must be defined at x = c. (ii) Te it x c f(x) must exist. (iii) Bot f(c) and x c f(x) must coincide. (ii) ( points) Wat is te derivative of a function f(x) at x = c (write it at as a it)? We define f (x) x c f(x) f(c) x c or equivalently, f (c) 0 f(x+) f(x). (iii) ( points) Give a geometric interpretation of te derivative of f(x) in terms of te curve y = f(x). For every x-value c, te derivative f (c) computes te slope of te tangent to te curve given by te equation y = f(x).. (8 points) In Figure 1 below you can see te grap of a function f(x) and te grap of its derivative f (x). Indicate wic one is wic and support your answer by giving at least two reasons. Te dotted curve is te grap of a some function f and te continuous curve is te grap of its derivative f (x). (i) Over tose intervals were te dotted curve is increasing (resp. decreasing), te continuous curve lies above (resp. below) te x-axis; namely, f is increasing (resp. decreasing) wenever f is positive (resp. negative). (ii) Wenever te dotted curve presents a local maximum or minimum value, te continuous curve crosses te x-axis (namely if f(x) is a local extreme point, ten f (x) = 0).

2 Figure 1: Question. ( points) Use te definition (it) to compute te derivative of te function f(x) = x + 5 Recall tat (a + b) = a + a b + ab + b. f (x) 0 f(x + ) f(x) 0 ((x + ) + 5) (x + 5) (x + x + x + ) + 5 x 5 0 6x + 6x + 0 = 6x 0 6x + 6x +. (8 points) Find te equation of te tangent line to te curve y = tan x at te point ( π, 1) Recall tat te slope of te tangent line to te curve given by te equation y = f(x) at te point (c, f(c)) is f (c), so tat te equation os te tangent line is given by y f(c) = f (c)(x c) In our case f(x) = tan x and c = π ; te derivative is given by f (x) = 1 ( π ) cos x, so f = 1 cos π = ( 1 ) = 1 so te equation of te tangent line is given by ( y 1 = x π ) = Page

3 Section B: Practical questions. 5. (16 points) Study te continuity of te function, x < 1 x f(x) = x, 1 x 1 x +x, x > 1 x by answering te following questions. (i) ( points) Wat is te domain of te function f (namely, at wat points is f defined)? Te function is defined over (, + ), except for te x-value x =. Note tat wen x > 1 (and clearly > 1), te function is given by x +x and te x denominator becomes zero at x =. Also observe tat wen x < 1, te function is given by f(x) =, and even toug te denominator becomes zero at x =, x tere is no discontinuity at x = since > 1. (ii) (6 points) Are tere any removable discontinuities? If so, wat is te it at tose points? (namely, if tere is a removable discontinuity at x = c, compute x c f(x)). Te discontinuity at x = is removable; indeed, te it is given by f(x) x = x + x x x = ( 1) = x (x 1)(x ) x x (x 1) (iii) (6 points) Are tere any jump discontinuities? If so, wat are te rigt-and and left-and its? Note tat as we approac te x-values x = 1 and x = 1 from te left and from te rigt, te function f as different expressions, so it is likely tat f presents jump discontinuities at tose x-values if te one-sided its differ. At x = 1 we ave f(x) x 1 x 1 x = 1, f(x) x = 1 x 1 + x 1 Since bot its coincide, we conclude tat f is continuous at x = 1. On te oter and, at x = 1 we ave f(x) x = 1, x 1 x 1 x + x f(x) x 1 + x 1 x Since bot its differ, f as a jump discontinuity at x = 1. = (iv) ( points) Are tere any vertical asymptotes? If so, wat are te rigt-and and left-and its ( or + )? Te function f as no vertical asymptotes. = 0 Page

4 6. ( points) Use te intermediate value teorem to sow tat te equation x 7x + 1x 8 = 0 as a solution in te interval [0,5] (Recall tat a solution of te equation f(x) = 0 is a number c suc tat f(c) = 0.) Te intermediate value teorem states tat if a function f is continuous over a closed interval [a, b] and f(a)f(b) < 0 (namely, te values of f at x = a and x = b ave opposite signs) ten tere exists some c inside [a, b] suc tat f(c) = 0. In plain words, if you draw a curve witout lifting your pen (i.e. a continuous curve), starting from below te x-axis and ending above te x-axis (or viceversa), ten you must ave crossed te x-axis at least once. In our case, te function f(x) = x 7x + 1x 8 is a polynomial, so it is clearly continuous and besides, f(0) = 8 < 0 and f(5) = = 1 > 0 Since te values of te function at te endpoints of te interval ave opposite signs, te Intermediate Value Teorem guarantees tat te equation must ave a solution inside [0, 5]. x 7x + 1x 8 = 0 7. ( points) Compute te it x x 1 + 8x x + x 1 + 8x x x 1 = x + = + 8 x x 1 + = = 8 = x 8. ( points) Compute te it x tan x x 0 sin x x tan x x 0 sin x x sin x cos x x 0 sin x x x 0 cos x = 0 1 = 0 Page

5 9. ( points) Compute te it x x + 7x + 1 x x + 7x + 1 = ( ) ( ) + 1 = = ( points) Compute te it (x + )(x x 6) x x + x + Evaluating at x = we obtain an indeterminate expression of te form 0. Tis means tat at least one monomial x + can be factored out from bot te 0 numerator and te denominator. Indeed, (x + )(x x 6) x x + x + (x + )(x + )(x ) x (x + ) x = 5 x 11. ( points) Compute te its x + 1 x + x 5x + 6, x + 1 x x 5x + 6 Evaluating at x = we obtain x+1, so te function f(x) = as a vertical 0 x 5x+6 asymptote at x =. In order to compute te left-and and rigt-and its, we factor te denominator as x 5x + 6 = (x )(x ) so tat and x + 1 f(x) x + x + (x )(x ) = + + x + 1 f(x) x x + (x )(x ) = + + Page 5

6 1. (0 points) Compute te derivatives of te following functions (you don t need to simplify your solution). (i) ( points) f(x) = (x + 1)(x + 1) Using te product rule we get f (x) = (x + 1) (x + 1) + (x + 1)(x + 1) = x (x + 1) + (x + 1) x (ii) ( points) f(x) = 5x +x 6 x 1 Using te quotient rule we get f (x) = (5x + x 6) (x 1) (5x + x 6)(x 1) (x 1) = (10x + )(x 1) (5x + x 6) (x 1) (iii) ( points) f(x) = sin x tan x f (x) = (sin x) tan x + sin x(tan x) = cos x tan x + sin x 1 cos x (iv) (8 points) f(x) = Note tat so tat x cos x+sin x x +1 (x cos x + sin x) = cos x x sin x + cos x = cos x x sin x f (x) = (x cos x + sin x) (x + 1) (x cos x + sin x)(x + 1) (x + 1) = ( cos x x sin x)(x + 1) (x cos x + sin x) x (x + 1) Page 6