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Can you tell whether I've taken this second derivative and determined its inflection points correctly?

$$f'(x) = (1-x)e^{ x-\frac{1}{2}x^2}$$ Now for the second derivative: $$f''(x)=(1-x)[e^{ x-\frac{1}{2}x^2}]' + (e^{ x-\frac{1}{2}x^2} [1-x]')$$ $$(1-x)(e^{x-\frac{1}{2}x^2} - e^{x-\frac{1}{2}x^2}x) + (-e^{x-\frac{1}{2}x^2})$$ Let $e^{ x-\frac{1}{2}x^2} = \psi$ $$(1-x)(\psi-\psi x) - \psi$$ $$\psi - 2\psi x + \psi x^2 - \psi$$ $$-2\psi x + \psi x^2 $$

Now determine inflection points: $$-2\psi x + \psi x^2 =0$$ $$\psi x(-2+x)=0$$ $$\psi x = 0 \text{ or } -2 +x = 0$$ $$e^{ x-\frac{1}{2}x^2}x = 0\text{ or } x = 2 $$ $$x = 0 \text{ or } x = 2$$

When I determine the inflection point I 'destroy' $e^{ x-\frac{1}{2}x^2}x $ by dividing on both sides by it. Is this a valid operation? Or should I have factored out only $e^{ x-\frac{1}{2}x^2}$ (without $x$)? $e^{ x-\frac{1}{2}x^2}= 0 $ doesn't have a solution and $-2x+x^2 = 0$ gives the desired inflection points (which are the same as I get using the method above ($x=0 \wedge x = 2$.)

Thank you,

Cro-Magnon

Michael Hardy
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Cro-Magnon
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  • I haven't checked the details but you should remember that not all points where the second derivative of a function is $0$ are inflection points. Since values of the exponential function are never $0$ you can certainly divide both sides by them. When the argument to the exponential function is real, then the value is necessarily positive, and that fact reduces the problem of finding where the second derivative changes signs to a problem not involving the exponential function. – Michael Hardy Apr 18 '16 at 13:48
  • Just above the line "Now determine inflection points" you got the sign of $2\psi x$ wrong. Oh, but then fixed it two lines later! – almagest Apr 18 '16 at 13:49
  • These 2 are. I checked. – Cro-Magnon Apr 18 '16 at 13:49
  • derivative is correct! – user190080 Apr 18 '16 at 13:51
  • @almagest Thanks I fixed that. – Cro-Magnon Apr 18 '16 at 13:52
  • $\ldots,{}$and now I've checked the details and it looks ok, except that you should explain why the two points you found are not only points where the second derivative is $0$, but also are points where the second derivative changes sign. – Michael Hardy Apr 18 '16 at 13:55
  • @michaelhardy Thanks for the tips and great explanation. – Cro-Magnon Apr 18 '16 at 14:04

1 Answers1

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So, you arrived at the step $e^{x-\frac{1}{2}x^{2}}x=0$ and you want to solve for $x$. Dividing both sides of this equation by $e^{x-\frac{1}{2}x^{2}}$ to conclude that $x=0$ is indeed a legal operation. This is because for all real numbers $r$, $e^{r}>0$. Therefore, for any $x$, $e^{x-\frac{1}{2}x^{2}}>0$. Since you are not dividing by $0$, you are okay.

ervx
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