Page 89 - Calculus for the Clueless, Calc II

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A. Graph y = xe = x/e ; y' = (1 - x)e = (1 - x)/e ; y" = (x-2)e = (x -2)/e :
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x
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1. x intercept(s) y = 0; top of fraction = 0; x = 0; point (0,0).
2. (0,0) is also the y intercept.
3. No vertical asymptotes, since the bottom of y is never 0.

4. Horizontal asymptote. This is different.

Asymptote is y = 0, but

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since x is negative and e gets very big as x goes to

We have a one-sided asymptote. Many exponentials have a one-sided asymptote.

5. Possible max, min y' = 0; top = 0; 1 - x = 0; x = 1; y = 1/e (1,1/e) y"(1) = (1 - 2)/e is negative, so (1,1/e) is a
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max.

6. Possible inflection points y" = 0. Again the top = 0; (x - 2) = 0; x = 2; y = 2/e . The point is (2,2/e ). It is an
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inflection point, since y"(2 ), y"(2 ) are different signs.
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Note
, of course, means equal by L'Hopital's rule. If you need to review curve sketching, look at Calc I Helper.
Now let's find the area of the right part.

B. Find the area of y = x e where x>0:
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The improper part is infinity.

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Integrate by parts: u = x and dv = e .