England_MAT_265_ONLINE_B_Summer_2019.jlalonzo.Section_4.1.pdf

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Jesse AlonzoEnglandMAT265ONLINEBSummer2019Assignment Section4.1 due 08/09/2019 at 11:59pm MST1.(1 point)The functionf(x) = (3x+7)e-6xhas one critical number.Find the critical number.f0(t) =0 whent=-1.2 andf0(t)does not exist whent=0.So the critical numbers are-1.2 and 0.Solution:Using the product rule yields:f0(x) =3e-6x-6(3x+7)e-6x= (-18x-39)e-6xthereforef0(x) =0 whenx=-136Answer(s) submitted:-13/6(correct)Correct Answers:-2.166666666666672.(1 point) Find all critical values for the functionf(r) =r10r2+9and then list them (separated by commas) in the box below.List of critical numbers:(1 point) Consider the functionf(x) =5-2x2,-5x1.The absolute maximum value isand this occurs atxequalsThe absolute minimum value isand this occurs atxequals
Answer(s) submitted:-3sqrt(10) / 10, 3sqrt(10) / 10(correct)
We havef0(t) =0 whent=-1.2 andf0(t)does not exist whent=0.So the critical numbers are-1.2 and 0.Answer(s) submitted:-6/50(correct)Correct Answers:-(2*3/5)04.(1 point) Consider the functionf(x) =5-2x2,-5x1.The absolute maximum value isand this occurs atxequalsThe absolute minimum value isand this occurs atxequals
Answer(s) submitted:
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