For t<0, analyze f(t)=t2∣t+1∣ in two regions.
For −1<t<0:
f(t)=t2t+1, so f′(t)=t3−(t+2)>0 (increasing).
For t<−1:
f(t)=t2−t−1, so f′(t)=t3t+2.
For −2<t<−1:
f′(t)<0 (decreasing).
For t<−2:
f′(t)>0 (increasing).
Largest decreasing interval is (−2,−1)=(2α,α), so α=−1.
For g(x)=2ln(x−2)−x2+4x+1:
g′(x)=x−22−2x+4=0 gives (x−2)2=1, so x=3.
Since g′′(x)=−(x−2)22−2<0, x=3 is maximum.
g(3)=2ln(1)−9+12+1=4