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What is the graph that best represents the function $(x+a)e^{-bx^2}$? a and b are positive constants Here are the options

My attempt:

At $x = 0, y=a$

Hence I eliminate the possibility of option d.

At $x=\pm \infty, y= 0$

Now I know the graph goes to zero at both ends.

At $x =-a, y=0$

This tells me that at a particular negative value of x the function goes to zero. Then for greater negative values of a, the function becomes negative and it subsides to the x-axis.

Hence I choose the option c.

Is this how generally is it done? or are there any method to break it down to make it easier to analyze the plot?

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  • $\begingroup$ The graph has one maximum and one minimum. You should find those next. $\endgroup$ – Paul Apr 18 '17 at 11:12
  • $\begingroup$ @Paul You mean I should take derivative of it? $\endgroup$ – rahul rj Apr 18 '17 at 11:16
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You should next find the maximum and minimum of the function. The derivative is $(-2bx^2-2abx+1)e^{-bx^2}$. Setting the derivative equal to zero you see that you get an extremum at the points $$x=\frac{2ab \pm \sqrt{4a^2b^2+8b}}{4b}$$. The negative root you can find is a minimum, and the positive root is a maximum. This should give you what you need to graph it.

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