Originally Posted by

**skeeter** $\displaystyle y = x - e^{-x}

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$\displaystyle y' = 1 + e^{-x}$

$\displaystyle 1 + e^{-x} = 3$

$\displaystyle e^{-x} = 2$

$\displaystyle -x = \ln(2)$

$\displaystyle x = -\ln(2)$

note that $\displaystyle -\ln(2) \ne \frac{1}{\ln(2)}$ ,

$\displaystyle -\ln(2) = \ln(2)^{-1} = \ln\left(\frac{1}{2}\right)$

to get the equation of the tangent line ...

at $\displaystyle x = -\ln(2)$

$\displaystyle y = -\ln(2) - e^{\ln(2)} = -[\ln(2) + 2]$

$\displaystyle y + [\ln(2) + 2] = 3[x + \ln(2)]$

$\displaystyle y = 3x + 2[\ln(2) - 1]$