if someone could show me all the steps to curve sketch y=3xe^-x that would be greatly appreciated!
My suggestions:
-Try to determine the domain and range of the function
-Find the - and - intercepts (if any)
-Find the derivative and determine where the function is increasing and decreasing
-Find the critical numbers of and use the first derivative test to find relative extrema
-Find second derivative and determine concavity and inflection points
-Sketch a few other points as needed
In this case, we have:
-Domain: is defined for all real , so our domain is the entire real line
-Intercepts:
so our only intercept is at .
-Increasing/decreasing and extrema:
Note that is defined everywhere. Setting , we have
and has one critical point, at . By testing on the intervals and , you should be able to determine that is increasing over the first interval and decreasing over the second, so this point is a relative maximum (and in fact it is an absolute maximum).
-Concavity and inflection points:
Differentiate:
So we have a possible inflection point at . Testing the intervals, you should find that (concave down) and (concave up). Thus is an inflection point.
Plot the appropriate points, and use the concavity and intervals of increasing/decreasing values to sketch the graph.
Hello, JMV!
The only intercept is the origin: (0, 0).Graph: .
The function is: .
Since , there are no vertical asymptotes.
Since , the horizontal asymptote is: . (x-axis)
First derivative: .
We see that: for , horizontal tangent (critical point).
. . For ... graph is rising (up to the critical point)
. . For ... graph is falling (down to the x-axis)
When . . . We have the point (1, 1.1)We can examine the second derivative.Code:| | o | * * | * * | * * - - - * - - - - - - - - - - - - - - - - - - - - * | * | | * | |
It simplifies to: .
. . For , concave down
. . At , inflection point at about (2, 0.8)
. . For , concave up
All of which agrees with my sketch . . . whew!