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HW 2 2–4 Let f (x) = (x − 1)2 − 1. (a) Explain why f has no inverse as is. (b) From here on, assume that the domain of f has been reduced to [1, ∞). Then the function f : [1, ∞) → [−1, ∞) is injective and its range is [−1, ∞) so that it has an inverse f −1 : [−1, ∞) → [1, ∞). Sketch the graph of f together with the graph of f −1 . (c) Find a formula for the function f −1 . (d) Verify that f and f −1 satisfy the inverse function identities (i) and (ii) (see 2.5). Solution: (a) Since f (2) = 0 = f (0), the two inputs 2 and 0 yield the same output 0, so f is not injective (and therefore it has no inverse). (b) The graph of f is a parabola that opens up (after expanding, the coefficient of the x2 term is positive). The vertex is (1, −1) (the smallest the expression can be is −1 and this occurs when x is 1). Reducing the domain to [1, ∞) retains only the right-hand side of the parabola. The graph of f −1 is obtained by reflecting through the 45◦ line y = x: Homework List JJ II J I Page 1 of 11 Back Home Page (c) f −1 (y) is the x for which f (x) = y, so we solve this last equation for x: f (x) = y ⇒ (x − 1)2 − 1 = y p ⇒ x−1=± y+1 p ⇒ x−1= y+1 p ⇒ x = y + 1 + 1, Homework List (1) where, at (1), the positive √ sign was chosen since x is in [1, ∞), implying that x−1 ≥ 0. Therefore, f −1 (y) =√ y + 1 + 1, or, switching to the more customary input variable x, we get f −1 (x) = x + 1 + 1. JJ II J I Page 2 of 11 Back (d) For x ∈ [1, ∞), we have f −1 (f (x)) = p f (x) + 1 + 1 = p ((x − 1)2 − 1) + 1 + 1 = |x − 1| + 1 = (x − 1) + 1 = x, where the penultimate equality uses that x ≥ 1. This gives (i). Home Page For x ∈ [−1, ∞), we have √ f (f −1 (x)) = (f −1 (x) − 1)2 − 1 = (( x + 1 + 1) − 1)2 − 1 = (x + 1) − 1 = x, giving (ii). 3–1 Simplify the following expressions: (a) 16−3/4 (b) 81/6 83/2 (c) log3 1 27 Homework List 4 ln 3 (d) e JJ II J I Solution: 1 1 (a) 16−3/4 = √ = 8 ( 4 16)3 √ 1 3 10 5 (b) 81/6 83/2 = 8 6 + 2 = 8 6 = 8 3 = ( 3 8)5 = 32 (c) log3 1 27 = −3, since 3−3 = 1 27 Page 3 of 11 Back 4 (d) e4 ln 3 = eln 3 = 34 = 81, using law of logarithms (iii) and then inverse property of logarithm (ii). Home Page 3–2 Sketch the graph of the equation y = 2 − ex+1 in steps: (a) Sketch the graph of y = ex+1 . (Hint: The graph is a shifted version of the graph of y = ex shown in 3.2.) (b) Sketch the graph of y = −ex+1 . (Hint: The graph is a reflected version of the graph of y = ex+1 .) (c) Sketch the graph of y = 2 − ex+1 . (Hint: The graph is a shifted version of the graph of y = −ex+1 .) Solution: (a) Homework List JJ II J I Page 4 of 11 Back Home Page (b) (c) Homework List JJ II J I Page 5 of 11 Back Home Page 3–3 2 (a) Solve e1−x = 2 for x. (b) Solve log5 (x − 4) + log5 x = 1 for x. Solution: (a) 2 e1−x = 2 2 ln e1−x = ln 2 1 − x2 = ln 2 (inverse prop. of log. (i)) Homework List 2 x = 1 − ln 2 √ x = ± 1 − ln 2. JJ II J I (b) log5 (x − 4) + log5 x = 1 log5 (x − 4)x = 1 log5 (x−4)x 5 (law of log. (i)) 1 =5 (x − 4)x = 5 2 x − 4x − 5 = 0 Page 6 of 11 (inverse prop. of log. (ii)) Back (x − 5)(x + 1) = 0 x−5=0 or x + 1 = 0 x = 5, −1. Home Page Since log5 (−1) is undefined, the solution x = −1 is extraneous and is therefore omitted. (In the solution, we applied the law of logarithms (i), which is not valid unless the factors are both positive, so this is how the extraneous solution entered in.) The solution is x = 5. 3–4 Show that any logarithm can be expressed in terms of natural logarithms by establishing the formula ln x loga x = . ln a Hint: Write y = loga x in the corresponding exponential form, apply ln to both sides, and solve for y. Homework List Solution: y = loga x ay = x ln ay = ln x y ln a = ln x ln x y= ln a JJ II J I (law of logs. (iii)) Page 7 of 11 Back 3–5 Suppose that the expression 21/3 has not yet been defined. Give an argument similar to those given at the end of Section 3.1 to show that 21/3 is forced by the laws of exponents √ 3 to equal 2. Home Page Solution: Using part (ii) in the laws of exponents, we have 1 (21/3 )3 = 2( 3 )(3) = 21 = 2. This says that 21/3 is a number that when cubed equals 2, which is to say 21/3 = 4–1 √ 3 2. Find the following without using a calculator. (Sketch the corresponding point P on the unit circle as well as the triangle you use to work out the answer, if appropriate): (a) cos 135◦ (b) sin(4π/3) (c) sin(−3π/2) Homework List (d) tan(π/6) JJ II J I Solution: Page 8 of 11 Back Home Page √ (a) cos 135◦ = −1/ 2 Homework List √ (b) sin(4π/3) = − 3/2 JJ II J I Page 9 of 11 Back Home Page (c) sin(−3π/2) = 1 triangle 30-60-90, we can use the (d) Since π/6 = 30◦ is acute and an angle of the known √ formula at the end of 4.2: tan π/6 = o/a = 1/ 3. 4–2 Homework List JJ II J I Sketch the graph of the equation y = 1 − sin(θ − π/4). Hint: Use the method of Exercise 3 – 2. Solution: The graph is determined in steps: Page 10 of 11 Back Home Page Homework List JJ II J I Page 11 of 11 Back Home Page