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Unformatted text preview: Solutions to Math 201 Final Exam from spring 2007 p. 1 of 16 (some of these problem solutions are out of order, in the interest of saving paper) 1. given equation: 1 2 ( x 1) 1 3 x = 4 both sides × 6: 6 · 1 2 ( x 1) 6 · 1 3 x = 6 · 4 multiply numbers to cancel denominators: 3( x 1) 2 x = 24 distribute: 3 x 3 2 x = 24 combine like terms: x 3 = 24 both sides +3: x = 27 . 2. given equation: x 2 = 7 x 5 2 both sides × 2x: 2 x · x 2 = 2 x · 7 x 2 x · 5 2 cancel denominators: 6 2 x · x 6 2 = 2 6 x · 7 6 x6 2 x · 5 6 2 clean up: x 2 = 14 5 x both sides 14 and +5 x : x 2 + 5 x 14 = 0 factor: ( x + 7)( x 2) = 0 answers: x = 7 or x = 2. So the sum of the two answers is 7 + 2 = 5 . 3. Note that this equation can be written in the form 2 u 2 + 3 u = 2, where the u represents the quantity x 1 . This is sometimes called a “quadratic form” equation. The strategy for solving them is to solve the equation for u first, and then when the uvalues are known, solve for x from there: equation: 2 u 2 + 3 u = 2 both sides 2: 2 u 2 + 3 u 2 = 0 factor: (2 u 1)( u + 2) = 0 finish solving for u : 2 u 1 = 0 or u + 2 = 0 = ⇒ u = 1 2 or u = 2. Replace u by x 1 : x 1 = 1 2 or x 1 = 2 rewrite negative exponent: 1 x = 1 2 or 1 x = 2 1 reciprocate (“flip”) both sides of each equation: x = 2 or x = 1 2 . Note: reciprocating both sides of an equation is only valid when you have (single fraction) = (single fraction). 4. This problem is just asking you to determine the smallest and largest allowable values that h can have, based on the given inequality. If we solve the inequality, the answers are revealed: inequality: 4 h 50 ≤ 30 rewrite without the : 30 ≤ 4 h 50 ≤ 30 all sides + 50: 20 ≤ 4 h ≤ 80 all sides ÷ 4: 5 ≤ h ≤ 20 . So minimum h value is 5 and maximum is 20 . 6. a) ( f + g )( x ) = f ( x ) + g ( x ) = x + 1 + √ x . b) ( fg )( x ) = f ( x ) g ( x ) = ( x + 1) p ( x ) . c) ( f ◦ g )( x ) = f ( g ( x )) = f ( √ x ) = √ x + 1 . d) ( g ◦ f )( x ) = g ( f ( x )) = g ( x + 1) = √ x + 1 . e) From part (d), ( g ◦ f )( x ) = √ x + 1 . The only restriction on x in this expression is caused by the squareroot. For a squareroot to be defined, the inside expression must be ≥ 0. So the domain here is x + 1 ≥ 0, which is the same as x ≥  1, which is written as the interval 1 , ∞ ) . 7. To find the formula for f 1 ( x ) when you are given the formula for f ( x ): 1) write the relation down with y as the output (rather than f ( x ), if the relation was given that way); 2) exchange x and y in the relation; 3) solve the new relation for y ; 4) replace y by f 1 ( x ). Here goes: original function: f ( x ) = 2 x 1 put in the y : y = 2 x 1 exchange x and y : x = 2 y 1 solve for y : both sides + 1: x + 1 = 2 y both sides ÷ 2: x + 1 2 = y replace y by f 1 ( x ): f 1 ( x ) = x + 1 2 . 5. This type of function is referred to in some texts as a “casedefined” function. The graph of such a function is madeThis type of function is referred to in some texts as a “casedefined” function....
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 Spring '08
 SMITH
 Math, Linear Programming, Optimization, feasible region, corner Point Principle

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