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**Macleef** Technicians working for the Ministry of Natural Resources have found that the amount of a pollutant in a certain river can be represented by $\displaystyle P(t) = 2t + \frac {1}{162t + 1}$, $\displaystyle 0 \leq t \leq 1$, where $\displaystyle t$ is the time (in years) since a clean-up campaign started. At what time was the pollutant at its lowest level?

$\displaystyle P'(t) = 2 - \frac {162}{(162t + 1)^2}$

$\displaystyle P'(t) = \frac {52 488t^2 - 160}{26244t^2 + 1}$

$\displaystyle 0 = 52 488t^2 - 160$

$\displaystyle 160 = 52 488t^2$

$\displaystyle \pm 12.64911064 = 52488t$

$\displaystyle 0.000240991 = t$

$\displaystyle P(0.000240991) = 0.963$

$\displaystyle P(0) = 0$

$\displaystyle P(1) = 2.006$

Therefore, the pollution in 0.000240991 year was at its lowest level of 0.963

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**Textbook answer: 0.049 years**

Did I take the first derivative wrong? I checked it and I don't see anything wrong with it. Please show me my mistake and fix it for me?