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?

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.