Asked by Janet
let f(x)=x^3(e^-x)
Answer using calculus, use graphing calculator only to check work.
a) Find the local and global extrema of f.
b) Find the intervals where f is increasing/decreasing.
c) Find the inflection points of f.
d) Find the intervals where f is concave up/down.
e) Find any asymptotic behavior (use limits to justify your answer)
Here's my work. Not sure I did it right so if someone could check. I don't know how to do e).
f'(x) = x^2(e^-x)(-x+3)=0
critical values are x=0 and x=3.
Not sure how to use these values to find part a.
f"(x)= x(e^-x)(-6x+x^2+6)
x=0, x=4.7321, x=1.2679 (find these with quadratic equation). These are the inflection points.
Increasing for x<3, until it hits x=3 which I think is the maximum (not sure if it's local or global extrema).
Decreasing for x>3, hit a local minimum at x=0 ( I'm not sure on these parts)
Concave up (0,1.2679), (4.7321, infinity)
Concave down (-infinity,0), (1.2679,4.7321)
I don't know how to do part e.
Answer using calculus, use graphing calculator only to check work.
a) Find the local and global extrema of f.
b) Find the intervals where f is increasing/decreasing.
c) Find the inflection points of f.
d) Find the intervals where f is concave up/down.
e) Find any asymptotic behavior (use limits to justify your answer)
Here's my work. Not sure I did it right so if someone could check. I don't know how to do e).
f'(x) = x^2(e^-x)(-x+3)=0
critical values are x=0 and x=3.
Not sure how to use these values to find part a.
f"(x)= x(e^-x)(-6x+x^2+6)
x=0, x=4.7321, x=1.2679 (find these with quadratic equation). These are the inflection points.
Increasing for x<3, until it hits x=3 which I think is the maximum (not sure if it's local or global extrema).
Decreasing for x>3, hit a local minimum at x=0 ( I'm not sure on these parts)
Concave up (0,1.2679), (4.7321, infinity)
Concave down (-infinity,0), (1.2679,4.7321)
I don't know how to do part e.
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