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| everlast746 Quote | Reply | | ahh! geometry problem posted on: 5/9/2006 9:06:44 PM if function g:x -> 8-3x find image of 5 and preimage of 0 i don't get it! this is the first question of the first section of the chapter |
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ahh! geometry problem
replied on: 5/10/2006 6:58:52 AM The image of g is defined as Image[g] = {g[x]| x \in D} where D is the domain. So the first question is simply what does the function g map 5 to? The preimage of a point y is defined as g^-1[y] = {x | g[x]=y} so the second question is what value(s) of x give g[x]= 0. As you have asked this question in the geometry section, I guess you want to use geometry. Well first plot the function g[x]. You will find it is a straight line with graident -3, which intercepts the x-axis at 8 and the y-axis at 8/3. After you have ploted this function you can simply read off the value for g[5], go 5 along the x-axis and then go up until you meet the line and then across to the y-axis and read off the value. For g[x] = 0 simply do the opposite. Go 0 up the y axis and then go along the x-axis until you meet the line, this is you preimage. Hope that it of some help |
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rajeev_sachdev
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ahh! geometry problem
replied on: 5/12/2006 2:09:31 PM g(x) -> 8-3x image of 5 <=> g(5) = 8- 3(5) => 8-15 => -7 if u wanna plot this pre-image on cartisian co-ordinate we have (x , g(x)) as (x,y) where y = g(x) so in this x = 5 and y = g(x) = -7 therefore points are ( 5,-7) pre-image of 0..this means that g(x) = 0 that is 8-3x = 0 => -3x = -8 => x = 8/3 if u wanna plot this pre-image on cartisian co-ordinate we have (x , g(x)) as (x,y) where y = g(x) therefore points are (8/3 , 0) Hope that satisfies u |
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