MATH 1A Midterm Practice September 29, 2014

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1 MATH A Midterm Practice September 9, 04 Name: Problem. (True/False) If a function f : R R is injective, ten f as an inverse. Solution: True. If f is injective, ten it as an inverse since tere does not exists x y suc tat f(x) = f(y). (True/False) If a function f is injective, ten it also must be surjective. Solution: False. Consider f : R R and f(x) = e x. Note tat it is injective, but it is not surjective since te range of f is (0, ). (True/False) If a function f is surjective, ten it also must be injective. Solution: False. Consider f : R R and f(x) = x 3 x. Note tat it is surjective since te range of f is R, but it is not injective since f(0) = f() = 0. (True/False) A function f must be eiter injective or surjective or bot. Solution: False. Consider f : R R and f(x) = x. It is bot not injective and not surjective since f() = f( ) = and te range of f is [0, ). (True/False) If f is increasing, ten f is injective. Solution: True. Since f is increasing, if x < y ten f(x) < f(y) and vice versa. Tus, if f(x) = f(y), ten x = y must be true. (True/False) If f : R R is injective, ten te range of f is te domain of f. Solution: True. Since f is injective, it as an inverse f and f can only invert values tat are in te range of f. (True/False) If f : R R is bijective, ten te range and domain of f must be te same.

2 Solution: False. Consider f(x) = ln(x). It is bot injective and surjective, terefore bijective. However, te domain of ln(x) is (0, ) wile te range of R. (True/False) A function f can be odd and even. Solution: True. However, tere is only function tat satisfies tis: f(0) = 0. (True/False) If f : R R is even, ten f is not injective. Solution: True. If f is even, ten f(x) = f( x), so f is not injective (it can never pass te orizontal line test). (True/False) If f is continuous at a and g is continuous at b and g(b) = a, ten f g = f(g(x)) is also continuous at b. Solution: True. f(g(x)) = f(g(x)) = f( g(x)) = f(a) = f(g(b)) x b g(x) a g(x) a (True/False) If f is continuous at a, ten f is also continuous at a. Solution: True. Since g(x) = x is continuous everywere, so g(f(x)) = f(x) is also continuous at a.

3 Problem. Find te domain/range of (a) ln(ln(x )) Solution: Let u = x and v = ln(u), ten we are simply left to solve te domain and range of ln(v). For te domain of ln(v), note tat we need v (0, ). Tis means tat v = ln(u) (0, ). Terefore, u = e v (, ). Lastly, u = x (, ). So, we solve x > x > x >, x <. Terefore, te domain is x (, ) (, ). For te range of ln(v), note tat te range of u = x is [, ). Note, we want to solve te range of v = ln(u) given tat u [, ). We see tat te range of v is R. Lastly, te range of ln(v) is also R. (b) tan(sin(e x )) Solution: Let u = e x and v = sin(u), ten we are just looking at tan(v). For te domain, note tat v = sin(u) [, ], so tan(v) is always defined. So, te domain is R. For te range, since we tat te range of u = e x is R. And te range of v = sin(u) is [, ]. Terefore, te range of tan(v) is [tan( ), tan()]. (c) e x+ x Solution: Let u = x + /x, ten we are just looking at e u. For te domain, note tat e u allows u R and x + /x is always in R unless x = 0. So, te domain is R {0}. For te range, we claim tat te range of u = x + /x is (, ] [, ). Tis is true because if x + /x = a, ten x + = ax x ax + = 0 x = a ± a 4 Note tat x as a solution wen a 4, so te range is any number wose square is 3

4 greater tan of equal to 4, wic is (, ] [, ). And te range of e u is terefore (0, e ] [e, ). Find te inverse of (d) y = x x + Solution: We solve for y wit x, y switced: x = y y + xy + x = y y(x ) = x y = x x (e) y = x x + wit domain x [0, ) Solution: We solve for y wit x, y switced: x = y y + xy + x = y y + xy + x = 0 So, by te quadratic formula, y = x ± x 4(x ) Since x [0, ), we take te positive root, y = x + x 8x + 4 Problem 3. Solve te following equations: (a) log (x) log x (e x ) = Solution: log (x) log x (e x ) = ln x ln e x ln ln x = x ln = x = ln (b) log (x) + log 3 (x) = 4

5 Solution: ln 6 = ln(x) ln() ln(3) (c) e x e x = log (x)+log 3 (x) = ln(x) ln + ln(x) ln 3 = ln(x)( ln() + + ln(3) ) = ln(x)ln() ln(3) ln() ln(3) = ln(x) = ln() ln(3) ln(6) ln() ln(3) x = e ln(6) Solution: Let u = e x, ten u u = 0 u = ± 5. Since u > 0, we must ave u = ( + 5)/ x = ln [ ( + 5)/ ] (d) sin (x) + cos(x) = Solution: Let u = cos(x), ten sin (x) = u. Tus, our equation becomes ( u )+ u = 0 u + u + = 0 u = ± = ± 3. Since u = cos(x) [, ], 4 our only solution is u = cos(x) = 3 x = cos ( 3 ) Note tat any π multiple added to x would also work, as cos(x) is periodic and x also works since cos(x) = cos( x). Te total solutions are: x = ± cos ( 3 ) + πk for any k Z (e) tan(x) + sec(x) = Solution: sin x + = cos x We can rewrite tis as: sin x cos x + cos x = Letting u = cos x and using sin (x) = u, we get: u = u + (u ) = u u u + = u u u = 0 u = 0, Tis gives cos x = 0, x = π, 0, π. However, plugging π, π into our equation gives us undefined values. So, te only valid solutions are x = πk for any k Z. 5

6 Problem 4. Solve te following its (witout LHopital s): (a) 0 Solution : 0 (b) = 0 + Solution : = + 0 = 0 + = + = 0 + ( + ) = 0 + ( + + ) = = 0 + ( + + ) Solution : = ( + )( + + ) 0 + ( + + ) Let u = +, ten u = and as 0, u. So, we rewrite as: u u u = u (c) 0 ( + ) /5 Solution: u u = u u u(u )(u + ) = u u(u + ) = Let u = ( + ) /5, ten u 5 = and as 0, u. So, we rewrite as: u u u 5 = u (d) x 4x + x Solution: u 4 + u 3 + u + u + = 5 ( x 4x + x = x 4x + x)( x 4x + + x) ( x 4x + + x) 6

7 x 4x + x = ( x 4x + + x) = 4 + /x ( 4/x + /x + ) = (e) sec(sin(x )) x Solution: Note tat sin(x ) [, ], so sec(sin(x )) [, sec()]. Terefore, x sec(sin(x )) x sec() x And since /x = sec()/x = 0, by te squeeze teorem, sec(sin(x )) x = 0 (f) sec( sin(e x )) x Solution: Note tat sin(e x ) [, ], but we see tat sec(π/) =. Tis means tat wenever sin(e x ) = π/, our function blows up: sec( sin(e x )) =. Terefore, tere is no it: sec( sin(e x )) = DNE x (g) x 3 (x ) x 3 Solution: Note tat te numerator is not zero wen you plug in 3. It is (3 ) =, so it is equivalent to: (x ) x 3 x 3 = DNE 7

8 Problem 5. Prove te following its (witout using it laws): (a) x x = 4 Solution : By te definition, we want to sow tat for any ɛ > 0 (condition ), we can find δ > 0 (condition ) suc tat 0 < x < δ x 4 < ɛ (condition 3) We first solve for δ: x 4 < ɛ x x + < ɛ x < ɛ/ x + Note we let δ (i just picked, but you can pick any oter constant), ten we know tat x < δ x x [, 3] Terefore, tis means tat x + [3, 5]. So, tis means tat ɛ/5 < ɛ/ x +. Terefore, δ = min(, ɛ/5). Now, we prove our statement. First, let ɛ > 0 be any positive number (condition ) and let δ = min(, ɛ/5) > 0 (condition ). We claim tat condition 3 is true: x < δ x 4 < ɛ Since x < δ = min(, ɛ/5), we know tat x < ɛ/5 and x <. If x <, ten x [, 3] x + [3, 5]. Terefore, x 4 = x x + < (ɛ/5)(5) = ɛ. So, x 4 < ɛ and tis concludes our proof. Solution (try to avoid): By te definition, we want to sow tat for any ɛ > 0 (condition ), we can find δ > 0 (condition ) suc tat 0 < x < δ x 4 < ɛ (condition 3) We first solve for δ: x 4 < ɛ x 4 ( ɛ, ɛ) x (4 ɛ, 4 + ɛ) x ( 4 ɛ, 4 + ɛ). NOTE: 4 ɛ could be problematic for ɛ > 4, so tis is wy tis proof is not encouraged...to make it fully accurate, we need to do casework on wen ɛ < 4 and ɛ > 4. 8

9 So, we ave x ( 4 ɛ, 4 + ɛ ), so we can take δ = min( 4 ɛ, 4 + ɛ ). Now, we prove our statement. First, let ɛ > 0 be any positive number (condition ) and let δ = min( 4 ɛ, 4 + ɛ ) (condition ). We claim tat condition 3 is true: x < δ x < δ and x > δ x < 4 + ɛ and x > ( 4 ɛ ) x < 4 + ɛ and x > 4 4 ɛ x < 4 + ɛ and x > ɛ + 4 ɛ > ɛ = 4 ɛ x 4 < ɛ And we are done. Note tat te algebra is more complicated...so tis proof is not recommended. (b) x 4x + = 5 Solution: I will not repeat te proof, but simply tell you ow to get δ. 4x + 5 < ɛ 4x 4 < ɛ 4 x x + < ɛ x < ɛ 4 x + We let δ, ten x < δ x [0, ] x + [, 3] Terefore, we pick δ = min(, ɛ/4). (c) x + = 0 Solution: By te definition, we want to sow tat for any ɛ > 0 (condition ), we can find N > 0 (condition ) suc tat x > N x + 0 < ɛ (condition 3) Since x + 0, we just solve x + < ɛ x + > ɛ x > ɛ x > ɛ Note tat we set N = ɛ, but N could be undefined for ɛ >. One way to correct tat is to just make N sligt bigger by letting N = ɛ. 9

10 Now, we prove our statement. First, let ɛ > 0 be any positive number (condition ) and let N = ɛ > 0 (condition ). We claim tat condition 3 is true: x > N x > ɛ x < ɛ x < ɛ Terefore, x + 0 = x + < < ɛ, as desired. So, we conclude. x (d) x 0 sin (x) = Solution: By te definition, we want to sow tat for any N > 0 (condition ), we can find δ > 0 (condition ) suc tat x 0 < δ sin (x) > N (condition 3) sin (x) > N sin (x) < /N sin(x) < and sin(x) > N N x < sin ( N ) and x > sin ( N ) = sin ( N ) x < sin (/ N) We let δ = sin (/ N). Now, we prove our statement. First, let N > 0 be any positive number (condition ) and let δ = sin (/ N) > 0 (condition ). We claim tat condition 3 is true: x < δ = sin (/ N) x < sin (/ N) and x > sin (/ N) sin(x) < / N and sin(x) > / N sin(x) > N and sin (x) > N So, we conclude. sin(x) < N sin(x) > N 0