$k(x) = e^{-\frac{x^2}{2}}$ on $[-1,2]$
I think the derivative of that is $ -x e^{-\frac{x^2}{2}}$. I don't know how to find zero from that equation.
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2Don't forget to check the endpoints of the interval. – Lythia Mar 12 '15 at 03:12
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the bracket is the intervals of the graph. – KPN123 Mar 12 '15 at 03:19
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Well, $x=0$ is a zero - there are no others – Dale M Mar 12 '15 at 03:20
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Oh ... and draw a picture - all will be revealed. – Dale M Mar 12 '15 at 03:21
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$k(x) = e^{-1/2*x^2}$ defined on [-1,2]$\subset R$. So $k'(x) = -xe^{-1/2*x^2}$ =0. Clearly k'(x)= 0 iff x= 0 since $e^{-1/2*x^2}\neq 0$ for all x in R. So the only non-endpoint critical point is at x= 0.We also have to check the endpoints on the bounded interval. At x=-1, k'(-1)= $\frac{1}{e^{1/2}}$. We now take the second derivative of k(x) to determine inflection:
$k''(x) = (x^{2})e^{-1/2*x^2}$=$\frac {x^{2}}{e^{\frac{1}{2}*x^2}}$
Unfortunately,at x=0, k''(0)=0. So the test is inconclusive. At k''(-1) > 0, so this is a local minimum at x=-1. At k''(2)>0,so this is also a local minimum at x = 2.
Now let's test k(x)'s behavior around the third critical point using the first derivative test. Let x =$frac{-1}{2}$. Then k'($frac{-1}{2}$) = $\frac{\frac{1}{2}}{e^{\frac{1}{32}}}$ > 0. Now consider x =1. Then k'(1)= $\frac{\frac{-1}{2}}{e^{\frac{1}{32}}}$ < 0. So x =0 is a local maximum for k(x). Therefore, the endpoints are local minimums and (0,1) is a local maximum. This means the function is increasing on [-1,0) and decreasing on [0,2],which is fairly clear from the formula. The graph also supports this:
Mathemagician1234
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