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Find the linear approximation of f(x)=\ln x at x=1 and use it to estimate ln 1.42
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interpret as "find the equation of the line tangent to \(y=\ln(x)\) at the point \((1,0)\)
you got that? take the derivative, plug in 1 to get your slope, then find the equation of the line using the point slope formula that is all
okay, so would the equation be y-0=1(x-1) ?
yeah or just \(y=x-1\)
how do you solve for ln 1.42? i plugged it into my calculator but its not the right answer
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replace \(x\) by \(1.42\) and get \(.42\)
it is kind of a lousy approximation but that is what you were asked for
ohh, thank you for the help!
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