if someone could show me all the steps to curve sketch y=3xe^x that would be greatly appreciated!(Doh)
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if someone could show me all the steps to curve sketch y=3xe^x that would be greatly appreciated!(Doh)
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).Quote:
Graph: .
The function is: .
Since , there are no vertical asymptotes.
Since , the horizontal asymptote is: . (xaxis)
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 xaxis)
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!