Solutions to Assignment #05 MATH = 2 p 5 2 (1) = 4 2p 5

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1 Solutions to Assignment # MATH Precalculus Section.7 (I) Complete Eercise #8 on p.. Solve > : If we solve the equalit, then we have two real roots b the Quadratic Formula. These are the critical values for the inequalit. q = ( ) () ( ) = p = p () : = :; :: This divides the number line into three intervals. We can choose test values in each interval. If the test value satis es the inequalit, then all the numbers in that interval. p (a) ; : =? ( ) ( ) > ) > : YES. p p (b) ; + : =? > : NO. (c) + p ; + : =? () > ) > : YES. Thus, the solution intervals are ; p ; + p ; + : Remember that since this was a strict inequalit, the critical values are NOT included. Also, we can check this with a graphing calculator or wolframalpha.com b looking at the graph and seeing where the graph is above the -ais ( > ) : We have = ; a parabola. We see that it is above the -ais on the end-intervals.x (II) Complete Eercises # on p.. Move everthing to the left side! : Factor b groups! = ( ) 9 ( ) = 9 ( ) = ( + ) ( ) ( ) = ) = ; ; :

2 There are four intervals. You ma want to use the Sign Chart. It looks like this: FACTORS ( ; ) ( ; ) (; ) (; +) ( + ) ( ) + + ( ) + f () + + So I ve arranged the factors in ascending order, in the sense that the associated critical values are in ascending order. We see that the factor ( + ) is negative in the rst interval when < the other intervals. and positive in Since f () = ( + ) ( ) ( ) ; we easil determine the sign of the product (or quotient, etc.). Thus, the solution intervals are ( ; ] ; [; +) : Since this is not a strict inequalit, we must include = and = in the solution. Sometimes this is easier than tring = = for the interval (; ) : = 8 : NO. We see that if we have rst power factors, the signs of the products/quotients alternate. This is a eas pattern to remember when it occurs. Let s check the graph. The curve is above the -ais on the two open intervals in the Sign Chart. (III) Complete Eercises # on p.. ( ) : There are onl two critical values. ( ) = ) = ; : (a) ( ; ): =? ( ) (( ) ) = : YES (b) (; ): =? () ( ) = : YES. (c) (; +): =? () ( ) = : NO. It turned out that the factor had no e ect on the product since it is alwas nonnegative. Since this is not a strict equalit, we DO include = and = in the solution.

3 The solution set is ( ; ]: The Sign Chart would be fairl simple. FACTORS ( ; ) (; ) (; +) ( ) + f () + Here s the graph: We note that the curve onl touches the -ais at = because = is a multiplicit zero! Recall that an even multiplicit will cause this to occur. (IV) Complete Eercises #8 on p.. We have ( + ) ( ) Our critical values are = ; in the numerator and = in the denominator. We know that can never be zero because f () would be unde ned. : FACTORS ( ; ) ( ; ) (; ) (; +) ( + ) ( ) + f () + + The solution intervals are [ ; ) ; [; +): We must include = and = since this is not a strict inequalit. Here s the graph: There is a vertical asmptote at = :

4 (V) Complete Eercises # on p If we move the left term to the right side, then we have f () = + + : I thought that this would be the easier function to graph. It is possible to go in the opposite direction also. We have vertical asmptotes at = and = : First, solve the equalit. We multipl through b the LCD, ( + ) ( all the denominators in one shot. ( + ) ( ) = + + ) : This eliminates ( + ) = ( ) + ( + ) ( ) + = + + = + 8 = = ( + ) ( ) = ; : So the critical values are = ; ; ; : Five intervals to test! (a) ( ; ): =? ( ) ( ) ( ) ( ) + + 8: YES! (b) ( ; ): =? ( ) ( ) 9 ( ) ( ) + + : NO. (c) ( ; ): =? : NO. (d) (; ): =? () + + : NO.

5 (e) (; +): =? All the choices are bad. () + + : YES! We know that = and = Here s m graph. must be int he solution since this is not a strict inequalit. I want the pieces where the graph is above the -ais. 7 8 The solution set is ( ; ) ; [ ; ) ; [; +) : Section. (VI) Complete Eercise #9 on p.. EXPLAIN our reasoning! Which graph is f () =? Recall that we can rewrite this as f () = = : This basic eponential function has a base a = = < : Thus, the curve must represent eponential deca as! +: It must be graph (a). [Similarl, for h () = ; we can let the elementar outer function be f () = and the inner function would be g () = : This would give us a horizontal shift of units to the right. Also, we could write h () = = = ( ) : This would give us vertical shrink b a factor of (=) : h () = (=) = =:X] It must be graph (b), because (VII) Complete Eercise # on p.. h () = + : Again, we want to use the elementar function f () = : We have h () = f ( ) + : This must have a hoirzontal shift units to the right and a vertical shift units up.

6 9 8 7 The reference point is (; ) : On the original graph, this was (; ) : (VIII) Complete Eercise # on p.. You will need a calculator or wolframalpha. f () = :e = ) f () = :e : = :978 : Your TI graphing calculator probabl said this: :978E. (IX) Complete Eercises # on p. 7. Compound interest. (a) n = : (b) n = : (c) n = : (d) n : (e) n = : (f) Continuous. A = + : = : = $8:: A = + : = : = $8:9: A = + : = : = $88:8: A = + : = $9:8: A = + : = $9:79: A = P e rt = e : = e : = $9:8: (X) Complete Eercises #8 on p. 7. Carbon-. We start with grams. Q = t=7 Ever time we go through one half-life of 7 ears, we lose about one-half of the original amount. (a) When t = ; we have Q () = (=) = grams.

7 (b) After ears, we have Q () = =7 := 7:8 grams. This is approimatel (=) =7 (c) Sketched in class. : = 78:% of the original amount. 7