Question 1065629: If f(x1, x2) = a + bx1 + cx2 + dx1x2, what is f(x1+k, x2) – f(x1, x2)?
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website! f(x1, x2) = a + bx1 + cx2 + dx1x2, what is f(x1+k, x2) – f(x1, x2)?
All those subscripts are confusing, so temporarily,
let x1=y, x2=z, then the problem becomes: f(y, z) = a + by + cz + dyz, what is f(y+k, z) - f(y, z)?We first find f(y+k, z) by replacing y by (y+k)
f(y+k, z) = a + b(y+k) + cz + d(y+k)z
Put the z by the d in the last term to make
it easier to distribute:
f(y+k, z) = a + b(y+k) + cz + dz(y+k)
Distribute:
f(y+k, z) = a + by + bk + cz + dzy + dzk
Get the last two terms in alphabetical order:
f(y+k, z) = a + by + bk + cz + dyz + dkz
Now we just put the equation for f(y, z) underneath
with like terms in line, and subtract:
f(y+k, z) = a + by + bk + cz + dyz + dkz
f(y, z) = a + by + cz + dyz
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f(y+k, z) - f(y, z) = bk + dkz
Four of the terms cancel, so we have
f(y+k,z) - f(y,z) = bk + dkz
Put back the confusing subscripts: y=x1, z=x2,
f(x1+k, x2) - f(x1, x2) = bk + dkx2
Edwin
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