University of Toronto Mississauga

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1 Surname: First Name: Student Number: Tutorial: University of Toronto Mississauga Mathematical and Computational Sciences MAT33Y5Y Term Test 2 Duration - 0 minutes No Aids Permitted This exam contains pages (including this cover page) and 5 problems. Check to see if any pages are missing and ensure that all required information at the top of this page has been filled in. No aids are permitted on this examination. Examples of illegal aids include, but are not ited to textbooks, notes, calculators, or any electronic device. Unless otherwise indicated, you are required to show your work on each problem on this exam. The following rules apply: Organize your work in a reasonably neat and coherent way, in the space provided. Work scattered all over the page without a clear ordering will receive very little credit. Mysterious or unsupported answers will not receive full credit. A correct answer, unsupported by calculations, explanation, or algebraic work, will receive no credit; an incorrect answer supported by substantially correct calculations and explanations might still receive partial credit. Problem Points Score Total: 30 If you need more space, use the back of the pages; clearly indicate when you have done this.

2 . Each question is worth mark. Indicate your answer in the given box. No part marks will be awarded for this question. (i) ( point) True or False: If A is an n m matrix and 0 is the n vector of all zeros, then the equation Ax = 0 is guaranteed to have a solution. Solution: The vector x = 0 is always a solution to this equation. True (ii) ( point) Let A be a 3 5-matrix, B a 7 5-matrix, and C a 3 7-matrix. Consider the expressions below, and write in the provided box all those expressions which are not defined. C T AB T, AC, AA T + CC T, BC, CB + A. Solution: The dimension of AC do not line up, as this is the product of (3 5)(3 7) matrices. Similarly, BC is the product of (7 5)(3 7) matrices. All other expressions are fine. AC, BC (iii) ( point) Determine the rank of the matrix A =. Solution: Every row of A is identical, hence row reducing results in a matrix with row echelon form We conclude that the rank of this matrix is. Page 2 of

3 (iv) ( point) Suppose A is an invertible matrix with inverse A. Which statement is not correct? A. The equation Ax = b has infinitely many solutions. B. A is a square matrix. C. The reduced row echelon form of A is the identity matrix. D. AA = A A Solution: The statement A is incorrect, as Ax = b has a single, unique solution x = A b. A (v) ( point) True or False: If x a f(x) = L then f(a) = L. Solution: This would be true if we knew f were continuous, but we don t know that s the case. Hence the statement is false. False (vi) ( point) Suppose x 4 f(x) = 7 and f(4) = 2. Which of the following statements is correct. A. f is continuous at x = 4. B. f has a removable discontinuity at x = 4. C. f is not defined at x = 4. D. f has a jump discontinuity at x = 7. E. No such function f can exist. Solution: The function is not continuous, since the value of the it differs from that of the function. On the other hand, since the two sided it exists, this means the discontinuity must be removable. B Page 3 of

4 (vii) ( point) Suppose f and g are continuous functions, and the chart below gives their values at particular points. x f(x) g(x) Determine the value of x [f(g(x)) g(f((x))] Solution: As both functions are continuous, the it is just f(g()) g(f()) = f(2) g( ) = 7 ( 4) =. (viii) ( point) True or False: If f is differentiable at c, then f is continuous at c. Solution: This is true, by a theorem in class. True (ix) ( point) Suppose that f is differentiable everywhere, and g(x) = f(xf(x)). Write down g (x) in terms of x, f, and f. Solution: Using the chain rule: g (x) = f (xf(x)) [ ] d dx xf(x) = f (xf(x)) [f(x) + xf (x)]. f (xf(x)) [f(x) + xf (x)] (x) ( point) Write down a function f which is continuous at all real numbers but is not differentiable at x = 2. Solution: There are many possible solutions. For example, f(x) = x 2 is a straightforward example. Variable Solutions. Page 4 of

5 2. Consider the matrix A = (i) (3 points) Find A if possible. If it is not possible, explain why not. Solution: We create the augmented matrix [A I 3 ] and row reduce to get R+R2 R R 0 0 +R 3 R R 2+R 3 R Since the matrix on the left is the identity, A is invertible and its inverse is the matrix on the right. Your Answer: (ii) (2 points) Let b = 2 and b 2 =. Define x as the solution to Ax = b and x 2 as the 3 solution to Ax 2 = b 2. What is 2x 3x 2? Solution: The vectors x and x 2 are easily determined using A found in part (a): x = A b = = 0 3 Thus x 2 = A b 2 = 2x 3x 2 = = 0 = Your Answer: 2 2 Page 5 of

6 3. Determine each it below. If the it does not exist, indicate whether it diverges to +,, or neither. x 3 3x + 4 (i) ( point) x 2 x + 2 Solution: This function is continuous at x = 2, so we simply evaluate to get x 3 3x + 4 = = 3 x 2 x (ii) ( point) 2x 2 4x 6 x x 2 + 3x + 2 Solution: We can factor an (x + ) term out of both the numerator and denominator: 2x 2 4x 6 x x 2 + 3x + 2 2(x + )(x 3) x (x + )(x + 2) 2(x 3) = 2( 4) = 8. x x (iii) ( point) x 2 x 4 x 2 6 Solution: The numerator is finite while the denominator is 0, indicating this is a vertical asymptote and that the it does not exist. The numerator is always non-negative, and the denominator is positive when x < 4, so x 2 x 4 x 2 6 =. Page 6 of

7 (iv) ( point) [ 4x4 + x 2x 2] x Solution: Multiplying by the conjugate we get [ 4x4 + x 2x 2] 4x 4 + x + 2x 2 x 4x4 + x + 2x = 2 (4x 4 + x) 4x 4 x 4x4 + x + 2x 2 x x 4x4 + x + 2x 2. From here we can see that the it will be zero, since the denominator is growing strictly faster than the numerator. Alternatively, the fastest growing term is the x 2 in the denominator, and we can renormalise to get x /x 4 + /x3 + 2 = 0 4 = 0. 0 (v) ( point) x 2 x 2 x 2 Solution: This it does not exist, but does not diverge to infinity. Indeed, in the it as we approach 2 from the left: while from the right x 2 x 2 x 2 2 x x 2 x 2 = x + 2 x 2 + x + 2 x 2 x 2 + x + 2 =. The one sided its fail to agree, so the two sided it does not exist. Does Not Exist Page 7 of

8 4. (5 points) Using either it definition of the derivative, find f (6) if f(x) = 2x x 2. Solution: There are two formulas for the derivative. In one case, we have Or alternatively, 2x f f(x) f(6) x 2 (6) 3 x 6 x 6 x 6 x 6 ( ) 2x 3(x 2) x 6 x 6 x 6 x 2 = 4. x 2 f f(6 + h) f(6) (6) h 0 h h 0 ( ) 2 + 2h 3(h + 4) h 0 h h + 4 h 0 h + 4 = 4 x 6 x 6 x 6 2(6+h) 6+h 2 3 h h 0 h ( ) 2x x 6 x 2 3 ( ) x + 6 x 2 h ( ) h h + 4 h 0 ( 2 + 2h ) h Page 8 of

9 5. (5 points) Let f(x) = x 4 e x 2 +3x. Find the equation of the tangent line to f at the point x =. Solution: At x = the value of the function is f() = 4 e +3 = e 2. The derivative requires both the product rule and chain rule: ( ) f (x) = 4x 3 e x 2 +3x + x 4 e x 2 +3x 2x + 3 2, x 2 + 3x so that f () = 4e 2 + 5e2 4 = 2e2 4. Using the point-slope formula, the equation of the tangent line is y e 2 = 2e2 (x ). 4 Alternatively, in y = mx + b form, the equation of the tangent line is y = 2e2 4 x 7e2 4. y e 2 = 2e2 (x ) 4 Page 9 of

10 This page is for additional work and will not be marked. Page 0 of

11 Formula Sheet Point-Slope Formula: y y 0 = m(x x 0 ) Product Rule: Quotient Rule: Chain Rule: d dx f(x)g(x) = f (x)g(x) + f(x)g (x) d f(x) dx g(x) = f (x)g(x) f(x)g (x) g(x) 2 d dx f(g(x)) = f (g(x))g (x) Page of