# week3c1 - x = 4 r 1 3 is a maximum Since y = x 1 x 4 is an...

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Week 3 Version 1 Class Time allowed: 15 minutes. 1. (4 marks) Let f ( x ) = x + 2 x + 7 . (a) Find the maximal domain S of f ( x ), and the range T of f ( x ). (b) Prove that f ( x ), considered as a function from S to T, has an inverse function g ( x ). Find a formula for g ( x ). 2. (2 marks) Find the x -coordinate of each stationary point of y = x 1 + x 4 . Determine the type of each stationary point. 3. (2 marks) Solve 1 3 x + 1 > 1 2 x - 3 .

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Week 3 Version 1 Class 4. (2 marks) Show that ln(sinh x ) < x - ln 2 for x > 0.
Week 3 Version 1 Class 1. (a) Maximal Domain: x R , x 6 = - 7 ; Range: y R , y 6 = 1 (b) f ( x ) = x + 2 x + 7 = x + 7 - 5 x + 7 = 1 - 5 x + 7 f 0 ( x ) = 0 + 5 ( x + 7) 2 > 0 for all x in the domain. Since the curve is monotonic increasing, it is one-to-one. [And since the codomain of f is its range, T, f ( x ) is onto.] Thus, it has an inverse function. f ( y ) = x 1 - 5 y + 7 = x 5 y + 7 = 1 - x 5 1 - x = y + 7 y = 5 1 - x - 7 So f - 1 ( x ) = g ( x ) = 5 1 - x - 7 2. y = x 1 + x 4 Using the quotient rule: dy dx = (1 + x 4 ) × 1 - x × (4 x 3 ) (1 + x 4 ) 2 = 1 - 3 x 4 (1 + x 4 ) 2 = 0 1 - 3 x 4 = 0 x 4 = 1 3 x = ± 4 r 1 3 . Stationary points occur when dy dx = 0, so there are two stationary points. At x = 0, y 0 = 1 1 > 0 , and at x = 1, y 0 = 1 - 3 (1 + 1) 2 < 0 The stationary point at

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Unformatted text preview: x = 4 r 1 3 is a maximum. Since y = x 1 + x 4 is an odd function, it has point symmetry about the origin. ∴ the stationary point at x =-4 r 1 3 is a minimum. Week 3 Version 1 Class 3. 1 3 x + 1 > 1 2 x-3 Critical Points: 3 x + 1 = 0 and 2 x-3 = 0 ⇒ x =-1 3 and x = 3 2 Equality: 1 3 x + 1 = 1 2 x-3 ⇒ 3 x + 1 = 2 x-3 ⇒ x =-4 Draw a number line and test points in each region: Figure 1: Number Line x =-5 : 1-14 > 1-13 : TRUE x =-1 : 1-2 > 1-5 : FALSE x = 0 : 1 1 > 1-3 : TRUE x = 2 : 1 7 > 1 1 : FALSE 1 3 x + 1 > 1 2 x-3 has solutions-1 3 < x < 3 2 and x <-4 4. For x > , e x-e-x < e x , since e-x > 0 for all x ∈ R . ⇒ ln( e x-e-x ) < x ⇒ ln( e x-e-x )-ln 2 < x-ln 2 ⇒ ln ± e x-e-x 2 ² < x-ln 2 ⇒ ln(sinh x ) < x-ln 2 as required....
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week3c1 - x = 4 r 1 3 is a maximum Since y = x 1 x 4 is an...

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