What is m if function y=(x^2+mx+1)e^x has extrema?



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embizze's profile pic

Posted on (Answer #2)

Find m such that `(x^2+mx+1)e^x` has extrema:

Since the function is continuous everywhere for any choice of m, extrema can only occur where the first derivative is zero.

`y=(x^2+mx+1)e^x` Use the product rule:


Setting `y'=0` we get:


Since `e^x!=0` we can use the quadratic formula:





So the extrema occur at x=-1 and x=-1-m.


Therefore m=-1-x


For example, if m=0 then the extrema occur only at x=-1

If m=-2 the extrema occur at x=-1,1

If m=-4 the extrema occur at x=-1,3

If m=4 the extrema occur at x=-1,-5

Some graphs:

Here is m=4 (it is hard to see that there is an extrema at x=-5 without zooming in):

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embizze's profile pic

Posted on (Reply #1)

I should have noted that if m=0 there are no extrema -- x=-1 is an inflection point.

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