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 May 23rd, 2010, 03:50 AM #1 Newbie   Joined: Apr 2010 Posts: 12 Thanks: 0 Second Derivative Test Let f(x) = (1 + x)^x Use the Second Derivative Test to determine whether f has a local maximum or a local minimum at x = 0. This question has gave me such a headache, as working out f''(x), is extremely long and I am thinking that there has to be some simple way to do it?
 May 23rd, 2010, 07:07 AM #2 Senior Member   Joined: Nov 2008 Posts: 265 Thanks: 0 Re: Second Derivative Test I believe the first derivative is $x[(1 + x)^x ^- ^1]$ Second derivative: $[(1+x)^x ^- ^1)] + x[(x-1)(1+x)^x ^- ^2]$ Sorry the exponents look weird, don't know how to use latex that well. They should be (x-1) or (x-2) Or I guess you could use logarithmic differentiation.
 May 23rd, 2010, 07:24 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,540 Thanks: 920 Math Focus: Elementary mathematics and beyond Re: Second Derivative Test $y\,=\,(x\,+\,1)^x$ $\ln{y}\,=\,x\ln{(x\,+\,1)}$ $\frac{y'}{y}\,=\,\ln{(x\,+\,1)}\,+\,\frac{x}{x \,+\,1}$ $y'\,=\,\left(\ln{(x\,+\,1)}\,+\,\frac{x}{x\,+\ ,1}\right)\left((x\,+\,1)^x\right)$ $y''\,=\,\left[\left(\ln{(x\,+\,1)}\,+\,\frac{x}{x\,+\,1}\right)( x\,+\,1)^x\right]\left(\ln{(x\,+\,1)}\,+\,\frac{x}{x\,+\,1}\right)\ ,+\,(x\,+\,1)^x\left(\frac{1}{x\,+\,1}\,+\,\frac{1 }{(x\,+\,1)^2}\right)$ $\text{At }x\,=\,0,\,y'#39;\,=\,2\,\Rightarrow\,\text{L ocal Minimum.}$
May 23rd, 2010, 07:50 AM   #4
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Re: Second Derivative Test

Quote:
 Originally Posted by mathman2 . . .the exponents look weird . . .
3^{x\,+\,1} gives $3^{x\,+\,1}$. Wrap the exponent in {}. The \, gives a bit of space between terms, signs etc.

 May 23rd, 2010, 08:19 AM #5 Senior Member   Joined: Nov 2008 Posts: 265 Thanks: 0 Re: Second Derivative Test Oh, thanks a lot greg1313. I had a feeling logarithmic differentiation would be more useful in this case.
 May 26th, 2010, 04:28 AM #6 Newbie   Joined: May 2010 Posts: 1 Thanks: 0 Re: Second Derivative Test there is first derivative test also

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