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f(130) = 64 f'(130) = 5 what is f(125)
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f(125) =14
f'(x) = c f'(130) = 5 c = 5 .. if I integrate it f(x) = cx + k c and k are constants f(130) = c*130 + k = 64 k = 64 - 130*5 f(125) = 5 *125 + 64 - 130*5
64-f(125)/(130-120) , then solve for f(125) ==> f(125)=14
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