MATH 126 Lecture Notes - Lecture 3: Horse Length

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fuchsiawombat825

15 Nov 2018

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Department

Mathematics

Course

MATH 126

Professor

Andrew D. Loveless

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Related Questions

Let P denote the vector space of all polynomials and f'(x)denote the derivative of all numbers in P. State true or false tothe following:

1. T: R^2 --> R^3 defined by T(x,y)=(?^2x,0, y/2) is a linear transformation

2. T: R^2 --> R^3 defined by T(x,y)=(x+y, x-y, xy) is alinear transformation

3. T: R^2 --> R^3 defined by T(x,y)=(1,0,0) is a lineartransformation

4. T: R^2 --> R^3 defined by T(x,y)= (y, x, y) is a lineartransformation

5. T: R^2 --> R^3 defined by T(x,y)= (x, x^2, x^3) is alinear transformation

6. T: R^2 --> R^3 defined by T(x,y)= (2x+3y, 3x+4y, 4x+5y) isa linear transformation

7. T: P --> P defined by T(f(x))= f(0) +f'(0)x + f''(0)x^2 isa linear transformation

8. T: P --> P defined by T(f(x))= f(1) + f'(2) + f''(3)x^2 isa linear transformation

9. T: P --> P defined by T(f(x))= f(x-3) is alinear transformation

10. T: P --> P defined by T(f(x))= f(x)-3 is a lineartransformation

tantoad80

[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) >

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MATH 126 Lecture Notes - Lecture 3: Horse Length

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