The function A(x) = 0.004x^3 - 0.05x^2 + 0.18x+ 0.05 approximates the blood alco

Question

The function A(x) = 0.004x^3 - 0.05x^2 + 0.18x+ 0.05 approximates the blood alcohol concentration in a person's bloodstream x hours after drinking 8 ounces of a hard liquor. The function only applies to the interval [0,5]. On what time intervals is the blood alcohol concentration increasing? On what intervals is it decreasing? Determine the interval(s), if any, on which the blood alcohol concentration is increasing. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. The blood alcohol concentration is increasing on the time interval(s) (Type your answer in interval notation. Use integers or decimals for any numbers in the expression. Round to the nearest hundredth as needed. Use a comma to separate answers as needed.) The blood alcohol concentration is never increasing. Determine the interval(s), if any, on which the blood alcohol concentration is decreasing. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. The blood alcohol concentration is decreasing on the time interval(s) (Type your answer in interval notation. Use integers or decimals for any numbers in the expression. Round to the nearest hundredth as needed. Use a comma to separate answers as needed.) The blood alcohol concentration is never decreasing.

Explanation / Answer

given A(x)=0.004x3-0.05x2+0.18x +0.05 , domain [0,5]

differentiate with respect to x

A'(x)=0.004*3x2-0.05*2x+0.18*1 +0

A'(x)=0.012x2-0.1x+0.18

concentration of alcohol is is increasing when A'(x)>0

0.012x2-0.1x+0.18>0

x<[0.1-((-0.1)2 -4*0.012*0.18)]/(2*0.012) ,x>[0.1+((-0.12) -4*0.012*0.18)]/(2*0.012)

x<2.63 ,x>5.70

interval is [0,2.63)

concentration of alcohol is is decreasing when A'(x)<0

0.012x2-0.1x+0.18<0

x>[0.1-((-0.1)2 -4*0.012*0.18)]/(2*0.012) ,x<[0.1+((-0.12) -4*0.012*0.18)]/(2*0.012)

x>2.63 ,x<5.70

interval is (2.63,5]

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