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I am attempting to find the vertical asymptotes, horizontal asymptotes, the local minimum and maximum, and the concavity of the function $f(x) = e^{(2x-x^2)}$

In order to find the vertical asymptotes, it is wherever f(x) is undefined, which I don't believe in anywhere. To find the horizontal asymptotes, I calculate the limit as x tends to infinity. Which is $0$.

I calculated the derivative. That is, $\dfrac{d }{dx}e^{(2x-x^2)} = e^{(2x-x^2)}$(2-2x)

I set it to zero and solve to get the local minimum and maximum.

I take the second derivative.

What does the second derivative tell me about the concavity? How is concavity even expressed in this graph for that matter?

James Graham
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    If the second derivative is (strictly) negative, then the function is concave. (It may be concave if the second derivative is zero, as in $x \mapsto x^4$, but not necessarily, as in $x \mapsto x^3$). If the second derivative is (strictly) negative, then the slope of the function is decreasing, which you can usually 'see' on graphs of these sorts of problems. – copper.hat May 02 '13 at 15:36

2 Answers2

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Hint: Note that $$ e^{2x-x^2}=e^{+1}e^{-(x-1)^2} $$ This implies that $ e^{2x-x^2} $ has graph similar to $e^{-x^2}$.

Elias Costa
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The second derivative is $f''(x) = 2 e^{2 x-x^2} \left(2 x^2-4 x+1\right)$ which is zero at $(2\pm\sqrt{2})/2$. Also, $2 e^{2 x-x^2}$ is always positive while $2 x^2-4 x+1$ is a parabola opening up and, therefore negative between the two roots while positive outside of the roots. Therefore, $f''$ itself is negative between the two roots and positive outside of the roots. This implies that the graph of $f(x)$ is concave down between the two and concave up outside them. Overall, the graph of $f(x)$ looks like so:

enter image description here

Mark McClure
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