Exam 1 Review Solutions

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1 Exam Review Solutions Please also review te old quizzes, and be sure tat you understand te omework problems. General notes: () Always give an algebraic reason for your answer (graps are not sufficient), () You may not use sortcut rules for derivatives (tese are covered in Capter 3- If you don t know wat I m talking about, don t worry about it). (3) We will usually use te a, version of te definition of te derivative- It s usually easier wit te algebra.. Finis te definition: (a) x a f(x) L if, for every ɛ > 0, tere is a δ > 0 suc tat: if 0 < x a < δ, ten f(x) L < ɛ (b) Te function f is continuous at x a if: x a f(x) f(a). Also, tis definition implies tree tings: (i) f(a) exists, (ii) x a f(x) exists, (iii) Items (i) and (ii) give te save value. (c) Te derivative of f at te point x a is: f (a) f(a + ) f(a) x. Find te domain: f(x) x To find te domain, we require tat: so we use a table to solve: (x + )(x ) ( + x)( x) 0 By te table, te domain is: (x + ) (x ) + ( + x) ( x) x < < x < < x < < x < x > x <, or < x, (wic is also were f(x) is continuous). (NOTE: It is easier to type a sign cart using te intervals under eac column- Wen you re writing by and, tat is not necessary- Just label te were te breaks are on te number line) 3. Find te domain: f(x) ln(x + x 3) Te domain of te natural log is te set of positive real numbers (not including zero). So we need to solve for x so tat x + x 3 > 0 (x + 3)(x ) > 0 Use a sign cart: So x < 3 or x >. (x + 3) + + (x ) + x < 3 3 < x < x >. Te formula for te equation of te tangent line to f(x) at x a is (use te point-slope form): Te point tat te line must go troug is (a, f(a)). Te slope of te tangent line is denoted by f (a). Put tese togeter (wit te point-slope form for a line) to get: 5. True or False, and give a sort reason: y f(a) f (a)(x a)

2 ( x (a) x x 8 ) ( ) ( ) x 8 x x x x x FALSE. Te it rules (Section.3) say tat we can only distribute te it troug a sum if te two its exist separately. In tis case, te two its do NOT exist separately. x + (b) x x 3 x (x + ) x (x 3) TRUE. Te it rules say we can do tis as long as te it in te numerator and denominator exist (and te it in te denominator is not zero). By tis rule, te it is (5/ ).5. By te way, tis is ow we sowed tat all rational functions are continuous on its domain. (c) If f is continuous and f() 3, f(), ten tere is an r so tat f(r) π. TRUE. Since 3 π (or f() π f()), and te function f is continuous, te Intermediate Value Teorem says tat tere is a c in te interval [, ] so tat f(c) π. (d) All functions are continuous on teir respective domains. FALSE. It is easy to construct a function tat is not continuous on its domain- We can use a piecewise defined function, like: { x + if x > f(x) 0 if x and so, wile f() 0, te it at x does not exist. (e) ln(x + 3) ln(x) + ln(3) FALSE. We do ave formulas for ln(ab) ln(a) + ln(b), but we do NOT ave formulas for ln(a + b). (f) sin (x) sin(x) FALSE. We reserve te exponent in functions to denote te inverse of te function, not te reciprocal. For example, ere te inverse sine function as domain x, but te domain of /sin(x) csc(x) are angles (excluding te points were sin(x) 0)- so te domains do not matc as well. (g) A vertical line intersects te grap of a function at most once. TRUE. Tis is te vertical line test to see if a grap is te grap of a function. Tat is, eac x can must ave a unique f(x). () If, wen taking a it of a rational function, we get 0, ten te it does not exist. 0 FALSE. If we get 0/0, we cannot say if te it exists or not (we must do more work). 6. For te function f(x) x, find and simplify te expression f(x + ) f(x ) 3 If f(x) x, ten f(x + ) (x + ) x + x +, and f(x ) (x ) x x +. Putting tese into te given expression, x + x + (x x + ) 3 6x 3 3 x 7. Sow tat tere must be at least one real solution to: x 5 x + We need to set tis up for te Intermediate Value Teorem, so we need to re-write te equation so tat f(x) 0. In tis case, we get: x 5 x 0 so tat f(x) x 5 x, wic is a polynomial, and terefore continuous for all x (continuity was a requirement of te IVT).

3 Now we begin substituting values of x into f until we observe a sign cange: f(0) f() f() so te IVT says tere is at least one c in [, ] so tat f(c) 0. Tis will be te solution to te original equation. 8. Solve for x: (a) x < x + 8 Rewrite, factor, and use a sign cart: x x 8 < 0 (x )(x + ) < 0 Te solution is te set of x so tat < x <. x + x x < < x < x > (b) e x Take te natural log of bot sides (or rewrite using logs): ln() x x ± ln() (We do note tat ln() > 0, so tis is a real solution). (c) ln(5 x) 3 Rewrite in exponential form: (d) ln(ln(x)) Rewrite in exponential form twice: We migt verify tat tis is a solution: e 3 5 x x 5 e 3 e ln(x) e e x ln(ln(e e )) ln(e ln(e)) ln(e ) ln(e) 9. If f(x) x 3, g(x) x, compute te expression for f g, g g (and simplify), g f. f g f(g(x)) f(/x) (/x) 3 x 3 g g x x g f g( x 3 ) x 3 ( 0. Use te ɛ, δ definition of te it to sow tat 3 5x ) x We begin wit our scratc work: Start wit f(x) L < ɛ, and we want to end wit x a < Expression in ɛ, wic will be δ. 3 5x < ɛ 5 5x < ɛ 5 ɛ x < ɛ x < 5 (Note tat a b b a ) 3

4 Now we give te actual proof:: Given any ɛ > 0, take δ ɛ 5. Ten:. Compute eac it algebraically (if it exists): 0 < x a < δ x < ɛ/5 5 x 5 < ɛ 5 5 x < ɛ 3 5 x < ɛ f(x) L < ɛ (a) x+ x x+ x x x + x + x + x x + x + x + x (b) x x (c) x x x 3 x 3 (d) (e) x x x x x x x x (x + ) x x + ( x + x + ) x x x 3/ (x + )(x )x (x + )x 6 x x x x x x x 3 3 (Be sure to ceck te numbers before doing any algebra!) ( x + x x) x + x + x x + x x x + x + x (x + x) x x + x + x x x + x + x x/x ( x + x + x)/x x +x x + + x + Remember, if x > 0, we can write x x, and if x < 0 (in te next problem), we will need to write x x. 3x + x x x 3x + x ( x (f) x ) 5x sin x 0 x 6 x 3 + x ( x )/x x 3 + x x x x 3 + x 3 x We can tink of tis as: x 0 x sin(a) I don t care wat A is, as long as it is going to infinity and not a number. If A were going to a number as x 0, ten te it could be found by substitution of x 0. Since A is going to infinity, we can use te Squeeze Teorem: x x sin(a) x Because te left and rigt sides of te inequality go to zero, so must te middle function. (Te answer is zero, by te squeeze teorem).

5 (g) x x x + 3x x x x + 3x x (x )(x + ) (x + 7)(x ) If you ad trouble factoring te denominator, be sure to use te fact tat we know tat (x ) is a factor- Tat is, if p(x) is a polynomial, and p(a) 0, ten (x a) is a factor for p(x). ( () x 0 + x ) x Before continuing, rewrite te function to get rid of te absolute values: x { 0 if x > 0 x x if x < 0 so te it as x 0 + is 0.. Find all vertical and orizontal asymptotes for x+3 x x 3 Vertical asymptotes for a fraction like tis will appear wen te denominator is zero AND te numerator is NOT zero. Factoring te denominator, we get x x 3 (x 3)(x + ). Te numerator is not zero at eiter x 3 or x, so tis function will ave two vertical asymptotes, x 3, x (RECALL tat x a is a vertical asymptote for f if f(x) ± ) x a ± For te orizontal asymptotes, we take te it as x ± : x x + 3 x x x x x 3 x x + 3 x x 3 x + 3 x Te lines y and y are orizontal asymptotes. + 3 x x 3x x x 3 x x + 3 x x 3x 3. Find all te values of a for wic f will be continuous for all real values. { x if x 3 (a) f(x) 3 ax if x > 3 Go troug all tree parts for continuity! Te only problem value(s) of x is at x 3 At tis point: f(3) 9 5, so f(3) exists. Te it as x 3 as to be found separately from te rigt and left: and 3 ax 3 3a x 3 + x 5 x 3 Terefore, for te it to exist, 3 3a 5, or a 8/3. At tis value of a, te it is 5. We see tat, if a 8/3, ten te first two numbers (f(3) and te it) are equal. If a 8/3, te function is continuous for every x. { x (b) f(x) if x a x if x > a Again we see tat, if x < a, f(x) x, wic is continuous. Also, if x > a, ten f(x) x is continuous. Te only problem point will be at x a. At tis point, we ceck te tree conditions for continuity: 5

6 f(a) a, so tis exists for all a. Te it must be found separately from te rigt and left: For te it to exist overall, we must ave: x a x a + x a x a a a a a 0 a ± Te it will exist if a ±. In tese cases, te its will be ±. If a ±, ten f(a) f( ± ) ±, wic makes te first two items equal. If a is eiter + or, we ave sown tat f is continuous at every x. { x 6 (c) f(x) x if x ± a if x ± Tere is no value of a tat will make tis function continuous, since te it does not exist at x (tat is, (x ) and (x + ) will not cancel out of te fraction. For example, x 6 x ± x ± (Te denominator is going to zero but te numerator is NOT).. Te displacement (signed distance) of an object moving in a straigt line is given by s(t) + t + t /, were t is in seconds. (a) Find te average velocity over te time period [, ]. s() s() Te average velocity is 6 3 (b) Find te instantaneous velocity at t. Te instantaneous velocity is s (): s( + ) s() ( + ( + ) + ( + ) /) (3/) ( 5 + ) ( ) 3 5. Find te equation of te tangent line to y 3x We need a point and a slope: at x 0. Te tangent line touces te function at x 0. Put tis into te formula for y to find te y coordinates: y. Te line terefore goes troug te point (0, ). Te slope of te tangent line is te derivative at 0. In tis case, f(0 + ) f(0) ( 3) 3 Te line troug (0, ) wit slope 6 is te tangent line: y 6(x 0) or y 6x ( 3)

7 6. Use te ɛ, δ definition of te it to sow tat: x 5 (7x 7) 8 Scratc work: start wit f(x) L < ɛ and simplify to x a : 7x 7 8 < ɛ We want x 5 Now we do te proof: Let ɛ > 0. Ten take δ ɛ 7, and: 7. Consider te function: 7x 35 < ɛ x 5 < ɛ 7 0 < x a < δ x 5 < ɛ/7 7x 35 < ɛ (7x 7) 8 < ɛ f(x) L < ɛ f(x) { x + 3, if x < x if x Te it of tis function at x does not exist (wy?). Draw a sketc of f(x) and sow ow te ɛ, δ definition of te it fails if we tink te it sould be L. 8. For eac function below, compute te derivative f (a) (use te definition of te derivative). (a) f(x) + x + (a + ) + a + a + a ( + (a + ) + + a) + (a + ) + a + (a + ) + + a + (a + ) + + a ( + (a + ) + + a) + a + a (b) g(x) x (a+) a a (a+) a (a+) a a a a (a + ) a a (a + ) a 3 7

8 (c) (x) x + x [a + + a + ] [a + a] a + a (d) f(x) 3 x (e) f(x) + x x a+ (a+) 3 (a+) 3 a ( + a + a + a + + a + a + + a ( 3 a 3 a ) 3 a 3 a 3 a 3 a ( 3 a + 3 a ) a a ) a + a ( a + + a) + a 3 a + 3 a 3 x + 3 a (3 a) 3 a) (3 a) 3/ (a + )(a ) a((a + ) ) a + a ((a + ) )(a ) ((a + ) )(a ) a (a ) 9. A space traveler is moving from left to rigt along te curve y x. Wen se suts off te engines, se will go off along te tangent line at tat point. At wat point sould se sut off te engines in order to reac te point (, 5)? (Hint: Label te unknown point on te grap of y x as (a, a )). An alternative way of stating tis problem: Find te equation(s) of te tangent lines to y x tat go troug te additional point (, 5). If we find tese values of x, since we re moving from left to rigt, we d coose te smaller x. Let a be te x coordinate we re looking for. Ten te line goes troug te points (a, a ) and (, 5). Tis says tat te slope of te line sould be: m a 5 a On te oter and, tis is a tangent line to x, so te slope of te tangent line at x a sould be te derivative, a. Putting tese togeter, a a 5 a Clear te fractions and simplify to get: a 8a Te solutions are a 3, a 5. We coose te smaller value, a 3, since we re moving from left to rigt. 0. Some algebra and trig: (a) (App D) Solve for x: tan(x) SOLUTION: Eiter sin(x) cos(x) or sin(x) cos(x). On te unit circle, tese are on te line y ±x, wic means te angles are π/, 3π/, 5π/, 7π/, and we could add multiples of π to tese (Alternatively: π/ + kπ, 3π/ + k/pi were k 0, ±, ±, etc. (b) (App D) Find all x so tat sin(x) SOLUTION: Consider sin(θ) on te unit circle- Points on te unit circle can be written as (cos(θ), sin(θ)), so anoter way to ask tis question is: Find te angle of all te y values on te unit circle less tan /. First we need to find te angles for wic sin(θ) /- Tose come from a triangle, θ π/6 and (on te unit circle), 5π/6. Terefore, te answer is: All angles on te unit circle except tose between π/6 and 5π/6. Anoter way to express tis: [0, π/6] [5π/6, π] (ten we could sift tese by multiples of π. 8

9 (c) (.3) Express F (x) / x + x as te composition of tree functions (you may not use te identity). SOLUTION: One way to do tis: ten F f(g((x))). (x) x + x g(x) x f(x) /x (d) (.6) Find te exact value: sin ( 3/) SOLUTION: Tink of te triangle. Tis is te angle θ for wic sin(θ) 3/, wic is π/3. Tere are no oter solutions, since te range of sin (x) was restricted to values between π/ and π/ (so tat te sine is invertible). (e) (.6) Simplify te expression: tan(sin (x)) If θ sin (x), ten sin(θ) x. Draw a rigt triangle wit tis relationsip - Make θ one of te angles, ten label te lengt of te side opposite as x, and te ypotenuse as. Terefore, te lengt of te side adjacent is x (by te Pytagorean Teorem). Now, te tangent of te angle θ is wat we want- We see (on te triangle we drew), tat te tangent is: x/ x. (f) (.6) Simplify te expression: sin(cos (x)) Similar to te last problem, we write θ cos (x), and ten cos(θ) x. Now draw a rigt triangle wit te appropriate relationsips (lengt of te side adjacent is x, te ypotenuse is, te lengt of te side opposite is x ) We see from te triangle tat te sine of θ is x /, or just x. (g) (.6) Solve for x: ln(x) < 3 SOLUTION: We could re-write tis as: < ln(x) ln(x ). Exponentiate bot sides: e < x If x > /e or if x < /e, we would get x > /e, owever negative values of x are not in te domain of ln(x) (wic is te original expression). Terefore, () (.6) Solve for x: e 7 x 6 SOLUTION: Take te log of bot sides (i) (.6) Find f and its domain if x > /e 7 x ln(6) x 7 ln(6) f(x) ln(e x 3) SOLUTION: Te domain of f is te set of x suc tat e x 3 > 0, or x > ln(3). Since te range of ln(x) is all real numbers, te range of f will be all reals. Terefore, te domain of te inverse will be all reals, and to find te inverse we switc x, y and solve for y: x ln(e y 3) e x e y 3 y ln(e x + 3) (j) (.6) If f(x) x 5 + x 3 + x, find f (3) and f(f ()) (Hint: Don t try to algebraically find f, try to eyeball it) SOLUTION: f (3) is te value of x so tat x 5 + x 3 + x 3. It looks like te answer is x. For te second expression, since f(f (x)) x, te answer is. (k) (.6) If f(x) + + 3x, find f (algebraically). SOLUTION: Switc x, y solve for y. You migt square bot sides first, ten we ll get: x y ( ) + 3y 3 x 9

10 . More True or False, and give a sort reason: (a) If a < b (and a > 0, b > 0), ten ln(a) < ln(b). TRUE: ln(x) is an increasing function. (b) If f(s) f(t), ten s t. FALSE: Only true if f is invertible and s, t are in te domain. (c) If f, g are functions, ten f g g f. FALSE: Easy to find an example- f(x) 3x, g(x) +x, ten f g 3(+x), wile g f +3x. (d) tan ( ) 3π/. FALSE: Te answer is π/. Te domain of tan(x) was restricted to angles between ±π/, so te inverse tangent cannot output angles bigger tan π/. (e) If x is any real number, ten x x. FALSE, if x < 0 (True, if we restrict x 0).. In te definition given in te text, f(x) L means tat for every ɛ > 0, tere is a number N so tat if x > N, ten f(x) L < ɛ. Using f(x) e x, give N and L if ɛ /e 5. Draw a sketc sowing tese values. SOLUTION: We sould see tat N 5; terefore, if x > 5, f(x) e Finis te definition: (a) A orizontal asymptote is: Te line y k, if (b) A vertical asymptote is: Te line x a, if f(x) k or f(x) k x f(x) ± x a +/ (c) A function f is one to one if: f(x ) f(x ) implies x x (or, if every y comes from at most one x). (d) A function f as even symmetry if: f( x) f(x) (e) A function f is decreasing if: a < b, ten f(a) > f(b). 0