MHF 4U Unit 7: Combining Functions May 9, 008. Review Solutions Use the ollowing unctions to answer questions 5, ( ) g( ), h( ) sin, w {(, ), (3, ), (4, 7)}, r, and l ) log ( ) + (, ) Determine: a) + w irst determine the corresponding points in () b) r Then add the y values. + w (, 4), (3,0), (4, ) h sin c) l log { } d) h e) g w substitute the y values in w, as, into g() to determine the y values o the composition. ( ) ( ) ) w r since the y values o r will become the values o w, we need to ind the values o in r that will result in the correct y values. g w {(, ), (3, 4), (4, 8) } w r {( ), ( 3 4, ), (, 7) }, 3 3 Page o 7.
) Graph h ( ) and w. Use your graphs to graph ( ) w h. w h() h( ) w 3) Determine: a) Dg l b) D l( r( ) ) c) R h( r( ) ) D ( r ( ))) { < or > 0 R} l, d) R w( l( ) ) R ( r ( )) { y y y R} h, R w ( l( ) ) {,, 7} 4) How many zeros does ( ) ( ) r have? Thus the product will have two zeros: 0 and MHF 4U Unit 7 Review - solutions Page o 7.
5) Will the product o ( ) and ( ) h be even, odd, or neither? () is an even unction and h () is an odd unction thus the product will be an odd unction. 6) Graphs o y () and y g() are given below. Sketch ( ) h ( ) on the grid above. g( ) () g() ( ) g( ) MHF 4U Unit 7 Review - solutions Page 3 o 7.
7) Given {(, 6), (, 8), (0, 5), (, 0), (, ) } and ( ) + a) graph g g, In order to subtract these unctions we need to ind the points on g() that match the values o the points on () g b) state the domain o g : D {,, 0,, } g (The overlap o the domains) g when g g() c) determine ( ) d) determine ( g( ) ) ( ) same as c) I meant g ( ( ) ) e) determine the domain o ( g( ) ) The other 3 values are not possible since + or all values. D g { 0, ± } MHF 4U Unit 7 Review - solutions Page 4 o 7.
8) I ( ) and g ( ), ind the value o or which ( + g)( ) 3. ± 9) Given ( ) + 3, determine a) ( ) b) c) a) ( ) 3 b) c) 0) Use composition to veriy that ( ) and + g( ) are inverses o each other. MHF 4U Unit 7 Review - solutions Page 5 o 7.
) Given ( ) sin and ( ) o. π g, describe the graph o g as a transormation o the graph 3 q, + 5 a) determine the intercept(s), asymptote(s), intervals o increase/decrease, and end behaviour o the unction, ) Given ( ) b) sketch the graph o the unction, MHF 4U Unit 7 Review - solutions Page 6 o 7.
c) deine an equation ( ) and an equation g ( ) such that ( ) q g ( ) ( ) d) justiy the properties you ound in a) by studying the properties o the unctions ( ) and ( ) D D D and g( ) 0 ; number o zeros number o zeros o, i in domain. q ( ) g g. both unctions have neither odd nor even symmetry and so their quotient also has neither symmetry. 3) Given ( ) sec a) the domain o g, where, π π and g( ) log determine, g g has VA at: the VA o () [ π/, 3π/ ] and the zeros o g() [ ] has a hole at the VA o g() [ 0 ] D π 3π,,, 0 < π R, g b) at most, the number o zeros or g, do you think this number is accurate? MHF 4U Unit 7 Review - solutions Page 7 o 7.
c) the domain o g D π 3π,, 0 < π R g, g has VA at: the VA o g() [ 0 ] and the zeros o () [none] g has a hole at the VA o () [ π/, 3π/ ] d) at most, the number o zeros or g, do you think this number is accurate? note: rom the graph is looks like the unction has zeros at the VA o (), but they are actually holes 4) Given ( ) 3 and g( ) tan a) the domain o g, where π π determine, π + nπ, π π, n Ζ R D g, MHF 4U Unit 7 Review - solutions Page 8 o 7.
b) the range o g Any real number multiplied by a number larger than zero will still be a real number. c) at most, the number o zeros or g, do you think this number is accurate? At most, the number o zeros o g number o zeros o + number o zeros o g. (Assuming there are no restrictions on the domain or common zeros, i.e.: double roots). Thus, at most, the number o zeros o g 5. All the zeros are within the domain and there are no common (double) zeros, thus this number is accurate. MHF 4U Unit 7 Review - solutions Page 9 o 7.
5) The plucked string o a guitar and the sound as it ades away can be represented by a damped sine t wave that has an equation o the orm y sin πt. Sketch a graph o the unctions or 0 t π. with graphing programme: y y sin π y sin π y MHF 4U Unit 7 Review - solutions Page 0 o 7.
6) Let S(t) represent the number o single adults in Canada in year t and M(t) represent the number o married adults in Canada in year t. Let E(t) represent the average amount spent on entertainment by a single adult and let N(t) represent the average amount spent on entertainment by a married adult. Using a combination o the unctions deined above come up with representations or the ollowing unctions: a) A(t), the number o Canadian adults in Canada in year t. A(t) S(t) + M(t), assuming that all adults are either single or married. b) B(t), the amount o money spent on entertainment by Canadian single adults in year t. B(t) (amount spent per single adult) * ( number o single adults) B(t) E(t) * S(t) c) C(t), the amount o money spent on entertainment by Canadian adults in year t. C(t) amount spent by single adults + amount spent by married adults. C(t) B(t) + M(t) * N(t) C(t) E(t) * S(t) + M(t)*N(t) 7) The change unction is deined as ( ) ( ) ( ) d. What is the meaning, in terms o, i is increasing is decreasing MHF 4U Unit 7 Review - solutions Page o 7.
8) Each o the ollowing graphs is a combination o two o the unctions: ( ), g( ) h ( ), j( ) cos, and one o the operations: addition, subtraction, multiplication, division, or each o the given graphs. State the equation and eplain how you know using key eatures o the unctions. Graph Equation and Eplanation y h( ) g( ) y The general shape o the curve is that o an eponential, thus it must be a combination with g(). The unction has a zero at 0. g(0), thus either was subtracted or the unction was multiplied by a unction with a zero at 0. (not divided because no asymptotes/holes). The y values o a cosine oscillate between - and thus the dierence between g() and () would also oscillate. Hence the unction must be a product. The zero is a turning point thus the unction is a product o g() and a unction with a double root, h(). y ( ) y cos j( ) The general shape o the curve is that o a sinusoidal, thus it must be a combination with j(). The unction has a zero at 0. j(0), thus either was subtracted or the unction was multiplied by a unction with a zero at 0. (not divided because no asymptotes/holes). The y values o a cosine oscillate between - and now they seem to be oscillating between a linear unction (when you join one set o etrema they orm the line y ; the other set orms the line y -. The unction is very likely a product o j() and (). Further supporting evidence: The zero at 0 is a single zero. The zeros o the unction are those o both j() and () j() is an even unction and () is an odd unction; the product o an even and an odd unctions is odd, such as this unction. MHF 4U Unit 7 Review - solutions Page o 7.
9) Suppose that some oil has been spilled in water and has ormed a circular oil slick. One minute ater the spill the radius o the slick is metres and 3 minutes ater the spill the radius is 6 metres. a) Epress the radius, r, o the spill as a unction o time, t, i the radius is increasing at a constant rate. b) Was the radius 0 at time t 0? c) Epress the circumerence, C, o the spill as a unction o time. d) Epress the area, A, o the spill as a unction o time. e) Determine the change unction or each o the radius, the circumerence and the area. What does it tell us about the spill? MHF 4U Unit 7 Review - solutions Page 3 o 7.
0) Find the unctions and g such that h ( t) ( g( ) ) h ( ) + b) a) ( ) 9 ( ) h c) h( ) sin( 3 + π ) 7 d) h ( ) 4 + + 4, given ( ) 5 and g () is a linear unction d) was challenging! ) Complete the ollowing table (determine the equation o the unction or draw the graph) MHF 4U Unit 7 Review - solutions Page 4 o 7.
a) Determine the domain o: In general, the domain is given by the overlapping values, quotient unctions have etra restrictions since the denominator can not equal zero. i) ( + k)( ) ii) ( p q)( ) iii) ( rs )( ) D { R } { > 0 R} D k, { > 0 R} D + k, v) ( m g)( ) D m { R } D g { R} zeros o g: -, 0, {, 0, R} D m g, b) Determine the range o: i) ( r + s)( ) { 0 R} { R} D p, D q, { 0, R} D p q, v) ( np )( ) D n { R} { 0 R} D p, D q { R} zeros o : -, { 0, ± R} D p q, ii) ( g)( ) {,, 3} s { 0,, } D r D D rs {, } r + s R r+s { 6,3} iii) ( h j)( ) {(, 6), (, 3) } { y y > 0 R} { y y > 0 R} R h, y R j, y A positive divided by a positive results in a positive R h j { y y > 0, y R} R { y y 4, y R} R q { y y R} The dierence o a quadratic and a cubic is a cubic R g iv) ( n )( ) { y y R} { y y 4 R} { y y R} R, y R n, y numbers -4 will be multiplied by, - and any number in between resulting in positive and negative # R n { y y R} MHF 4U Unit 7 Review - solutions Page 5 o 7.
c) Algebraically, determine whether the ollowing are even, odd or neither i) ( p )( ) ii) ( jn )( ) ( p )( ) ( 4) ( p )( ) even ) ( ) 4 ( ) ( 4) ( p)( ) ( p)( ) ( p)( ) This supports our conclusions: the product o two even unctions is even. iii) ( )( ) mm ( ) ( )( ) sin sin ) ( mm)( ) sin( ) sin( ) ( mm)( ) sin( ) sin( ) ( mm)( ) sin( ) sin( ) ( mm ( mm)( ) ( mm)( ) even This supports our conclusions: the product o two odd unctions is even. d) Determine all the zeros or the unction ( jn)( ) cos() ( jn)( ) ( jn)( ) ( jn)( ) ( jn)( ) not not even odd ( ) ( jn)( ) ( jn)( ) cos( ) cos() ( cos() ) cos() ( jn)( ) ( jn)( ) This supports our conclusions: the product o a neither with an even unction is neither. iv) ( m p)( ) ( ) ( m p)( ) sin ( m p)( ) sin odd ( ) ( ) ( ) ( m p)( ) sin ( m p)( ) ( m p)( ) This supports our conclusions: the quotient o one odd unction with one even unction is odd. i) ( g )( ) zeros o : -, zeros o g : -, 0, zeros o g : - (DR), 0, (DR) iii) ( l )( ) zeros o l :, > 0 zeros o : -, zeros o lg :, ii) ( m g)( ) zeros o m : 0, ± π, ± π, in general: ± nπ zeros o g : -, 0, zeros o m g : ± π, ± π, in general ± nπ NOT - (VA), (VA), 0 (hole) iv) ( h q)( ) zeros o h : none zeros o q : none zeros o m g : none NOT - (hole) MHF 4U Unit 7 Review - solutions Page 6 o 7.
e) Determine the average rate o change in the interval [, 3] or the unctions i) () avg (3) () RoC 3 5 ( 3) 3 4 ii) g () avg g(3) g() RoC 3 5 ( 3) 3 9 iii) m () avg m(3) m() RoC 3 0.9975 0.4794 ɺ 3 ɺ 0.5905 iv) p () avg p(3) p() RoC 3 9 3 4 9 ) Determine the average rate o change in the interval [, 3] or the unctions i) ( + g)( ) avg RoC avg 4 + 9 3 RoC + avg RoC + g ii) ( gm )( ) This is NOT equal to the product o the two avg RoC! 3 gm ( 4) ( sin ) ( )( ) avg ( gm) (3) ( gm) () RoC 3 g(3) m(3) g() m() 3 4.965 (.438) ɺ 3 ɺ 8.004 iii) ( m p)( ) avg RoC m p avg RoC m 4 ɺ 0.5905 9 ɺ 0.7035 avg RoC p iv) ( p)( ) This is NOT equal to the quotient o the two avg RoC! ( 4) ( ) ( p)( ) ( p) (3) ( p) avg RoC () 3 (3) p(3) () p() 3 45 ( 3) 3 4 MHF 4U Unit 7 Review - solutions Page 7 o 7.