Determine the intervals where the function f(t)=x^2e^-x is increasing and where
ID: 3346316 • Letter: D
Question
Determine the intervals where the function f(t)=x^2e^-x is increasing and where it is decreasing. Please show calculus and leave answer in interval notation.
Determine the intervals where the function f(x)=xe^x is concae up and where it is concave down. Please show calculus and leave answer in interval notation.
The percentage of alcohol in a person's bloodstream t hr after drinking 8 fluid oz of whiskey is given by A(t)=.23te^-.4t . How fast is the percentage of alcohol in a person's bloodstream changing after 4 hours? The first part of the question was what is the percentage of alcohol in a person's bloodstream after 4 hours. I just plugged in the 4 for t and came up with .1857. Please show the calculus on the part I asked about so I can understand. Thanks!
Explanation / Answer
Determine the intervals where the function f(t)=x^2e^-x is increasing and where it is decreasing. Please show calculus and leave answer in interval notation.
F'[X]=[E^(-X)][-X^2+2X]=0...FOR OPTIMUM .....X=0 AND 2 ARE THE 2 POINTS .....
F'[X] IS NEGATIVE FOR X<0............POSITIVE FOR X>0..............NEGATIVE FOR X>2.....SO .......CONCAVE UP AROUND X=0....(-INFINITY , 2]......AND..............
..CONCAVE DOWN AROUND X=2 .....[ 0 , + INFINITY )
Determine the intervals where the function f(x)=xe^x is concae up and where it is concave down. Please show calculus and leave answer in interval notation.
F'[X]=[E^X][X+1]=0..FOR OPTIMUM.........X=-1 IS THE POINT ....
F'[X] IS NEGATIVE FOR X<-1 ....POSITIVE FOR X>-1.............
SO CONCAVE DOWN AROUND X= -1 .......(-INFINITY , INFINITY)
The percentage of alcohol in a person's bloodstream t hr after drinking 8 fluid oz of whiskey is given by A(t)=.23te^-.4t . How fast is the percentage of alcohol in a person's bloodstream changing after 4 hours? The first part of the question was what is the percentage of alcohol in a person's bloodstream after 4 hours. I just plugged in the 4 for t and came up with .1857. Please show the calculus on the part I asked about so I can understand. Thanks!
RATE OF CHANGE = DA/DT = [0.23][E^(-0.4T)][-0.4T + 1] ...SO ..T =4 WE GET ...
RATE OF CHANGE = - 0.0279 ...DECREASING AT A RATE OF 0.0279 ...ANSWER
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