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Absolute max/min for f(x)=ln(x+2)+1/x
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f'(x) = 1/x+2 - 1/x^2 = 0 1/x+2 = 1/x^2 x+2 = x^2 x^2 -x -2 = 0 (x-2)(x+1) = 0 x = 2, -1 -1 outside domain so disregard look at f' value for x less than 2 to determine concavity f'(1) = 1/3 - 1 = -2/3 slope neg so concave up gives min min f(x) is x=2
wow thank you
your welcome actual answer is f(2), just plug in 2 for x
Okay thanks agian
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