Max Extrema and Point of Inflection

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To find the maximum extrema and point of inflection for the function g(t) = 1 + (2+t)e^(-t), the first derivative g'(t) = -e^(-t)(1+t) indicates a critical point at t = -1, which is a maximum. The second derivative g''(t) = (t)e^(-t) reveals a point of inflection at t = 0. To determine the coordinates, substituting t = -1 into g(t) yields the maximum point (-1, 1 + e), and substituting t = 0 gives the point of inflection (0, 3). Understanding how to derive these points from the original function is crucial for accurate analysis.
kari82
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1. find the extremas and points of inflections



2. g(t)= 1+(2+t)e^(-t)



3. So i know you need to find g(t)' and g(t)''

g(t)'= -e^(-t)(1+t)
g(t)''= (t)e^(-t)

my critical point is t=-1 (max) and my point of inflection is t=o

How do I get to max extrema being (-1, 1+e) and point of inflection (0,3)? Thank you very much!
 
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You've only given the x-coordinates of your critical point and point of inflection. The actual point is (t, g(t)). In other words you need to plug the values for t you have into the original function to get the corresponding y-coordinate.
 
Thank you so much! I knew it wasnt hard... I just couldn't figure it out. Thanks again!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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