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Calculus1 8 Online
OpenStudy (anonymous):

Can someone tell me if I did this correctly? I was to locate absolute extrema of the function on the closed interval. h(x)=-x^2+3x-5 [-2,1]

OpenStudy (anonymous):

If I differentiate it would be h'(x)= -2x+3 Then setting equal to zero would be x=3/2 which is a critical number Then, I evaluated at -2,1 and 3/2 and found that the minimum would be -15 and the maximum would be -1/2

OpenStudy (anonymous):

Perfect

OpenStudy (anonymous):

thank you so much. I felt like i was missing something thats why i was making sure

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