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2.
When racer 1 has double racer 2’s speed their functions will be equal. But racer 1 is now picking up speed. So his function is now… j (x) = 10sin((pi/5)(x))+40 +x Because every x twentieth of the track he picks up x speed so your adding the x twentieths of the track to his speed And racer 1 will be trying to double racer 2’s speed so racer 2’s function that has double the speed and half the range is… k(x) =5 sin((pi/5)(x))+ 70 (5 before the sine would stay the same because it would be doubled and halved) . Since their functions will be equal we can make an equation out of this… j (x) = k(x) 10sin((pi/5)(x))+40 = 5sin((pi/5)(x))+ 70
3.
We then try to solve for x… j (x) = g(x) 10sin((pi/5)(x))+40 +x = 5sin((pi/5)(x))+ 70 10sin((pi/5)(x))+40 +x - ( 5sin((pi/5)(x))+ 70) = 0 5 sin((pi/5)(x)) + x – 30 = 0 5 sin((pi/5)(x)) + x = 30 5 sin((pi/5)(x)) = 30 – x Any way that you manipulate it you are going to get an x within the argument of sine. But from here it is solvable.
4.
5 sin((pi/5)(x)) = 30 – x Take the reciprocal of both sides 1/(5 sin((pi/5)(x) ) ) = 1/(30 – x) The outputs of 1 /(5 sin((pi/5)(x) ) ) are + when 0<x<5 + a10; a E Real The outputs of 1 /(5 sin((pi/5)(x) ) ) are - when 5<x<10 + a10; a E Real Domain: {x|x cannot equal 5a; aEI; xER} Range: {y|y<=-0.2, y=>0.2;yER} From what we know about reciprocals 1/(30 – x) will have invariant points at y = –1 and 1. Setting it equal to those outputs… 1/(30 – x) = 1 1/(30 – x) = -1 x = 29 x = 31 We have invariant points at (29,1) and (31,-1)
As this reciprocal function approaches the x axis it will never cross the other reciprocal graph because…
the values closest to 0 for our reciprocal sine function are 0.2 and –0.2.
These values do not occur in the same range as when the function of 1/(30-x) has values further or equidistant from the x-axis. This is because in the ranges that 1/(30-x) has outputs greater than or equal to 0.2 (25<=0<30) and smaller than or equal to –0.2 (30<x<=35 ) our reciprocal sine graphs output’s are of opposite sign or have an asymptote and therefore they will never intersect (y=-0.2 @ x=35, y=0.2 @ x=25 for our 1/(30-x) function.). (Shown on next slide)
These functions don’t intersect and yet when you don’t take the reciprocals of both sides they do. We know they do because they both have the domain (-infinity,infinity), 30-x has a range of (-infinity,infinity) and that the sine function has a range of [-5,5]. Since this is true, for 30-x to leave the positives and enter the negatives at atleast one point they must intersect.
7.
The reciprocal function 1/(30-x) has only 1 input that it cannot use and that input (x = 30) must be where the other graphs meet, since they intersect and reciprocal graphs didn’t. If you input it into the question for x it is true. 5sin((pi/5)(30)) = 30 – 30 5sin(6pi) = 0 5(0) = 0 0=0 30 is how many twentieths of the track it will take. So it will take him one and a half laps at this ever increasing pace to double his opponents speed.
8.
30 – x (blue) and 5sin((pi/5)(x)) (green) Intersection at 30 supports our answer.
9.
10sin((pi/5)(x))+70(black) and 10sin((pi/5)(x))+40+x (red) Intersection at 30 supports our answer.
10.
There is a shorter way to do this though. At one point we had 5 sin((pi/5)(x)) = 30 – x. Now if you recognize that if x was any multiple of 10 that you would get sin(multiples of 2pi), which no matter what is 0 and 5 times 0 is 0. 30 is a multiple of 10 and would give you zero on the other side of the equation which would make both sides of the equation equal and 30 a valid answer.
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