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📰 For \(v > 1\), \(g'(v) < 0\), so \(g(v)\) is decreasing on \((1, 2]\). Hence, maximum at \(v \to 1^+\)? But let’s check value at endpoint. 📰 Wait — since derivative is negative, function decreases from \(v = 1^+\) to \(v = 2\), so maximum occurs as \(v \to 1^+\), but \(v > 1\), and we want finite maximum. 📰 But observe: as \(v \to 1^+\), g(v) → ∞? Let’s test: 📰 Breaking News Pollo Loco Stock Soars Find Out Why This Burger Giant Is Unstoppable 4907879 📰 Phantom Hourglass Shock The Hidden Feat That Ruined And Fixed Your Gameplay 2638103 📰 T Mobile Home Internet 2909019 📰 Mp Stocktwits Explosive Updates The Inside Look No Investor Can Ignore 5924458 📰 Broken Cast Iron Rusted Out Heres How To Restore It Super Fast 2136501 📰 Classic 90S Gems You Need To Watchthese Movies Are Packed With Nostalgia And Heart 3372853 📰 Abbott Elementary Season 5 5291494 📰 The Ultimate Guide To Creating Sql Tables That Boost Your Database Performance 7845569 📰 Bryson Dechambeau Driver 27560 📰 Charles Schwab Cut Dramaticallywhat This Means For Your Investments Is Scandalous 9830234 📰 Best Bank Account To Open 4596187 📰 This Little Trick Turned Every Pedal Stroke Into Pure Rewards On Velo 3267754 📰 All Avatar Movies 2404166 📰 Fire Red Pokmon Gameplay Simplified Complete Walkthrough Youve Been Searching For 4555195 📰 You Wont Believe The Star Power Behind Joshua Jacksons Biggest Movie Tv Moments 461374