MATH114 Lecture Notes - Lecture 4: Pythagorean Theorem, Quadratic Equation, Farad

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orangeturtle356

14 Sep 2018

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University of Alberta

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Mathematics

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MATH114

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Alsaody,Seidon

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MATH114 Lecture 3: Absolute Values Properties, Intro to Trig

MATH114 Lecture Notes - Lecture 4: Pythagorean Theorem, Quadratic Equation, Farad

MATH114 Lecture Notes - Lecture 5: Compound Interest, Exponential Function, Logarithm

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As x is a variable, we need to consider different cases. So, we divide into: x x-2 x < 0. From the table above we can see how the value of x and x-2 varies. For x 2, therefore |x| = x so solve x > 3(x - 2) 2x<6 x < 3. For 0 x <2 |x| = x so solve x > - 3(x-2) similarly as before. For x < 0 |x| = -x so solve -x > - 3(x-2) similarly as before. Put all the solutions together at home as an exercise. Use pythagorean theorem: x2 + (cid:4672)(cid:2869)(cid:2873)(cid:4673)(cid:2870)= 1 x2 = 1 - (cid:4672)(cid:2869)(cid:2873)(cid:4673)(cid:2870) = (cid:2870)(cid:2872)(cid:2870)(cid:2873) x > 0, = (cid:2870)(cid:2872)(cid:2870)(cid:2873)= (cid:2870) (cid:2874)(cid:2873) Therefore cos ( ) = (cid:2870) (cid:2874)(cid:2873) and tan=(cid:2929)i(cid:2924)c(cid:2925)(cid:2929)=(cid:2869)/(cid:2873)2 (cid:3122)(cid:3121) = (cid:2869)(cid:2870) (cid:2874) Trigonometric functions: cos2(v) + sin2(v) = 1, pythagorean theorem https://slideplayer. com/slide/9319341/ From the functions above, we get: sin(u-v) = sin(u+(-v)) = sin(u)cos(-v) + cos(u)sin(-v)

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indigohound134

Hello there. I just had one question dealing with binary operationsin abstract algebra. I need to see if this operation has anidentity element for all x in R. The operation is as follows:x*y=|x-y| This is what I did. I basically broke it up since thereis an absolute value. x*e=|x-e| = x-e if x> 0 or -(x-e) if xe,then x*e= x-e. We need to find e for x*e=x. So x-e=x and e=0. Nowmy first question is this: 1) If x>0, then why for thisequation, x-e=x, can x be values that are less than 0. It seemslike a contradiction to me. Is e=0 only valid for x>0? Thenafter that I tried out x*0= |x|= x if x>0 and -x if x<0.However, is it just a coincidence that it is x if x>0 and alsowe chose x*0 to be x-0 for x>0? I mean, it seems like either way( whether you use the |x| way or the (x-e) way, you get x as ananswer). My next question is that if we choose x<0 for the abovex*0=..., then doesn't e change? I am confused about this. I mean inone case we have |x|= -x if x<0, but if x<0, then e changes,doesn't it? Please just help me answer these questions and guide mestep by step through this problem. Thank you.

periwinklejackal701

[1] 1. (a) Solve 2 – 31 < 7. Solution: We have – 7<-3 < 7 which gives -4 0 on (-0, 1] U [2,). Next, we need to see where Vr2 – 3r +2 -1 > 0. If r2 - 3.c + 2 – 1 > 0, then V.22 – 3.0 +2 > . Observe that this is definitely true if r < 0. If r > 0, then squar- ing both sides we get 22 - 3.+2 > 1 2 > 3. V A VICO Thus, the domain of f is (-00, ). [2] (c) Find the exact value of tan(sec-14). Solution: Let y = sec-14. Then, sec y = 4. From the diagram we get that tan(sec-? 4) = tan y = 15 [2] (d) Find all r such that sin(2.x) = COS I. Solution: We have 2 sin r cos r = COS I, SO 0 = 2 sin Cos I -Cos I = cos .r (2 sin c - 1). Hence, sin(2x) = cost when cos x = 0, or when sin r = Thus, r= +ik, kez, I= +2nk, ke Z, or r= +2nk, kez [2] (e) State the precise mathematical definition of lim f(x) = 0. Solution: For every e > 0 there exists a 8 >0 such that if 0 < x-al <, then f(c) >

MATH114 Lecture Notes - Lecture 4: Pythagorean Theorem, Quadratic Equation, Farad

MATH114 Lecture 3: Absolute Values Properties, Intro to Trig

MATH114 Lecture Notes - Lecture 4: Pythagorean Theorem, Quadratic Equation, Farad

MATH114 Lecture Notes - Lecture 5: Compound Interest, Exponential Function, Logarithm

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