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f(x)=3x^3+4x^2-x+C and f(2)=1 what is the value of c
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put in '2' for 'x' and set the whole thing equal to '1'... then find C
ok im trying it
so I get fx=3(8)+4(4)-x+c
Replace all x with 2 So you should have f(2)=3(8)+4(4)-2+c which can be simplified more so
Then set that =1 and solve for c
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you would get : fx=3(8)+4(4)-2+c = 1 now c is easy to find,right?
im trying to work it
c = 1 - 24 - 16 + 2 which is: -37
C=-37
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