MATH 155A FALL 13 PRACTICE MIDTERM 1 SOLUTIONS. needs to be non-zero, thus x 1. Also 1 +

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1 MATH 55A FALL 3 PRACTICE MIDTERM SOLUTIONS Question Find te domain of te following functions (a) f(x) = x3 5 x +x 6 (b) g(x) = x+ + x+ (c) f(x) = 5 x + x 0 (a) We need x + x 6 = (x + 3)(x ) 0 Hence Dom(f) = {x R x 3, } (b) Te denominator of x+ needs to be non-zero, tus x Also + x+ + x + = x + x + 0 x + 0, or yet x Hence Dom(g) = {x R x, } 0, wat gives (c) We ave Dom( 5 x) = {x R x 5} and Dom( x 0) = {x R x 0} Since te domain of te sum is te intersection of te domains, we ave ie, tis f is not well defined Dom(f) = {x R x 5} {x R x 0} =, Question An electricity company carges its customers a base rate of $0 a mont, plust 5 cents per kilowatt-our (kw) for te first 00 kw and 7 cents per kw for all usage over 00 kw Express te montly cost E as a function of te amount x of electricity used E(x) = { x, 0 x (x 00), x > 00 Question 3 At te surface of te ocean, te water pressure is te same as te air pressure above te water, 5 lb/in Below te surface, te water pressure increases by 434 lb/in for every 0 ft of descent Express te water pressure as a function of te dept below te ocean surface Let be te dept below te ocean surface and P te pressure Ten P = ,

2 MATH 55A FALL 3 wit P measured in lb/in and in ft Question 4 Compute te values of te following trigonometric expressions (a) sin 5π 6 (b) tan 9π 4 + cos( π 6 ) (c) sec 4π 3 (a) 9π (b) tan cos( π) = 3 (c) sec 4π = 6 3 cos 4π 3 = tan(4π + 3π) = tan 3π =, cos( π) = cos( π) = = = Question 5 Prove te following formulas (a) sin x sin y = sin(x + y) sin(x y) (b) cos θ = +cos(θ) so (a) We ave (a) We ave so sin(x ± y) = sin x cos y ± sin y cos x, sin(x + y) sin(x y) = (sin x cos y + sin y cos x)(sin x cos y sin y cos x) = sin x cos y sin y cos x = sin x( sin y) sin y( sin x) = sin x sin x sin y sin y + sin y sin x = sin x sin y cos(θ + ξ) = cos θ cos ξ sin θ sin ξ, cos(θ) = cos(θ + θ) = cos θ sin θ Adding te above formula wit = cos θ + sin θ yields te result Question 6 Find all solutions to te following trigonometric equations (a) cos x = 0 (b) tan x = (c) + cos x = 3 cos x So tan 9π 4 +

3 or (a) We ave cos x =, wic gives x = π 3 + πk, x = 5π 3 MATH 55A FALL πk (b) We ave tan x = or tan x =, wic gives x = π 4 + πk, x = 3π 4 (c) Use cos x = cos x (see question 5) to write + cos x = 3 cos x, cos x 3 cos x + = 0 Tis is a quadratic equation for cos x Te quadratic formula gives cos x = 3 ± 9 8, 4 so cos x = or cos x =, wat gives x = 0 + πk, x = π 3 + πk, x = 5π 3 + πk + πk Question 7 Evaluate te following its, sowing tat te it does not exist wen tat is te case (a) x + x 3 x + 3 x 3 (b) x x x3 + (c) x 3x (d) x 0 x3 + x sin π x (e) tan x x π (f) ( x 0 + x ) x ( (g) x 0 x ) x () x π sin(x + sin x)

4 4 MATH 55A FALL 3 (i) x 7 + x (a) Te function is defined at 3, so te it is 5 6 (b) Write for x, x 3 x = x3 x x + x + = ( x + )(x 3 ) x = ( x + )(x )(x + x + ) x = ( x + )(x + x + ) Tis expression is defined at x = and te it is terefore 3 = 6 (c) Te function is defined and continuous at x =, so x3 + x 3x = x x 3 + 3x = 7 4 (d) Notice tat x 0 x3 + x = 0, x 0 ( x 3 + x ) = 0, and sin π x Hence x 3 + x x 3 + x sin π x x 3 + x From te squeeze teorem we terefore conclude tat x3 + x sin π x = 0 x 0 (e) Since tan x wen x π and tan x wen x π +, we conclude tat tan x as x π (f) Because x = x for x > 0 and te it is from te rigt, we can remove te absolute value and ten = = 0 Hence te it is equal to zero x x x x (g) Because x = x for x < 0, and te it is from te left, if we remove te absolute value: = + = for x < 0 Hence x x x x x ( x 0 x ) = x x 0 x = We computed te it from te rigt above and found zero Hence te it does not exist as te its from te rigt and left do not agree

5 () Since sin x is continuous MATH 55A FALL 3 5 sin(x + sin x) = sin( + sin x) = sin(π + sin π) = sin(π + 0) = sin π = 0 x π x π x π (i) Again by continuity x 7 + x = + x 7 x = + 7 Question 8 Let x if x <, 3 if x =, g(x) = x if < x, x 3 if x > Evaluate or explain wy te it does not exist (a) g(x) x (b) x g(x) (c) x g(x) (d) g(x) x + (e) x g(x) (a) (b) (c) (d) (e) does not exist Question 9 For te function g of te previous question, indicate te values of x for wic g is not continuous Te function is discontinuous at x = and x = Question 0 domain Explain wy te following functions are continuous at every point in teir

6 6 MATH 55A FALL 3 (a) f(x) = sin x x + (b) f(x) = tan x 4 x (c) f(x) = sin(cos(sin x)) (a) sin x and x + are continuous Te quotient of continuous functions is continuous wenever te denominator does not vanis (b) tan x is continuous were it is defined, x is continuous for x 0 and 4 x is continuous Te composition 4 x is terefore continuous were it is defined, since te composition of continuous functions is continuous Te quotient tan x 4 x is terefore continuous on its domain (c) Composition of continuous functions is continuous Question Let f(x) = x3 8 Can you define a new function, g(x), wic agrees wit f(x) x 4 on te domain of f(x) and is continuous at x =? Wat value sould f() ave if we want to define it as a continuous function at x =? Notice tat for x ± x 3 8 x 4 = (x )(x + x + 4) (x )(x + ) = x + x + 4 x + Define g by te same expression as f for x ±, and put g() = = 3 Also, define f() = 3 Question Using te ε, δ definition of a it, sow tat (a) (3 4 x) = 5 x 0 5 (b) x x = 0

7 MATH 55A FALL 3 7 (a) Write x ( 5) = x = (x 0 + 0) = 4 x 0 5 Given ε > 0 we can ten coose δ = 5 4 ε (b) For given ε > 0, we want, for x > 6, We can terefore coose δ = ε x 0 = 8 x ( 6) < ε Question 3 Using ε, δ arguments, prove tat te function f(x) = point on its domain x+ is continuous at every Fix a Given ε > 0, we want x + < ε, a + or x a a + x + < ε If δ is suc tat δ < a+, ten x + > a+ wenever x a < δ Tus So to get or equivalently, x a a + x + < x a a + a + x a a + < ε, x a < a + ε, = x a a + we can coose δ = a+ ε Terefore, if δ = min{ a+, a+ ε} we conclude tat x a < δ implies x + < ε, a + wat sows tat te result x+ is continuous at a Since a is an arbitrary point in te domain, we ave Question 4 Using te definition of derivative, compute f (x) (a) f(x) = x

8 8 MATH 55A FALL 3 (b) f(x) = x 3 + x So (a) Write (b) Write Hence f(x + ) f(x) f(x + ) f(x) f (x) = 0 f(x + ) f(x) (x+) x 3+(x+) 3+x = (x + ) x = x + x + x x + = = x + = 0 (x + ) = x = = (3 + x)( x ) ( x)(3 + x + ) (3 + x + )(3 + x) = (3 + x)( x) (3 + x) ( x)(3 + x) ( x) (3 + x + )(3 + x) = (3 + x) ( x) (3 + x + )(3 + x) (3 + x) ( x) 7 = = (3 + x + )(3 + x) (3 + x + )(3 + x) f (x) = 0 f(x + ) f(x) 7 = 0 (3 + x + )(3 + x) = 7 (3 + x) URL: ttp://wwwdisconzinet/teacing/mat55a-fall-3/mat55a-fall-3tml