From previous parts of this question, we know that E ( t ) = 1 48 t ( t 2 − 24 t + 192 ) {\displaystyle E(t)={\frac {1}{48}}t(t^{2}-24t+192)} . The plot shows this function. Note that when E ′ ( t ) = 0 , t = 8 {\displaystyle E^{\prime }(t)=0,t=8} and E ′ ′ ( 8 ) = 0. {\displaystyle E^{\prime \prime }(8)=0.} Thus 8 {\displaystyle 8} is the only extremal point, and it is an inflection point. The average error rate is E ( t ) t = 1 48 ( t 2 − 24 t + 192 ) = 1 48 ( ( t − 12 ) 2 + 48 ) = 1 48 ( ( t − 12 ) 2 ) + 1 {\displaystyle {\frac {E(t)}{t}}={\frac {1}{48}}(t^{2}-24t+192)={\frac {1}{48}}((t-12)^{2}+48)={\frac {1}{48}}((t-12)^{2})+1} .
Answer: the minimal value is 1 {\displaystyle \color {blue}1} .
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