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Alan

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 the roots of 3x^2 -2x-5=0 are alpha and beta find the equation whose roots are alpha^2 and beta^2 (Posted for sacha)

So alpha = 5/3 and beta = -1
To get the (quadratic) equation whose roots are alpha2 and beta2 we construct:
(x - alpha2)*(x-beta2) = 0 or:
x2 -(alpha2+beta2)*x + alpha2*beta2 = 0 lee mas ..
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Alan  22 abr 2014
 #5
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...I didn't think that would work properly - they are not usually that compatible.

I copy everything between, and including, begin{document} and end{document}

I haven't used any esoteric features so I don't know what the limitations of this are!

29 abr 2014
 #3
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I haven't used either of begin{align} or begin{equarray}, I'm afraid.  

I suspect that my problems are more likely to be of my own making in the main as I'm still very much a beginner with LaTeX.  I find it easier to use Texmaker and then copy and paste into the LaTex Formula box than write it directly here.   

29 abr 2014
 #4
avatar+33661 
+3

1.  Look at a graph of your function between the limits of 0 and 2:

x-2lnx+1)    

You can see that there is a single minimum in the range.

2. To find the turning points on a curve you need to find the places where the tangents are zero.  A zero tangent means that a line through that point is horizontal.  You can see from the curve that a tangent line at the minimum of the curve will be horizontal.

3. The tangent line to a curve at any point has the same gradient as the curve at that point.  The gradient is obtained by differentiating the function with respect to x. So if the curve is described by f(x) = x - 2ln(x+1) you have to differentiate this to find the gradient.  Differentiating the first term, x, with respect to x is just 1.  Differentiating the second term with respect to x gives 2/(x+1).  Overall then the gradient of the curve at any point is given by gradient(x) = 1-2/(x+1).

4.  We want the point where this gradient is zero, so we set 1-2/(x+1) = 0 and find the value of x that makes this true.  Here we can see that x = 1 is required since 1-2/(1+1) is indeed zero.

5.  The value of f(x) at this point is obtained by putting x = 1 into the expression for f(x) -  this will give us the value of the minimum of f(x):  1-2ln(1+1)  or 1-2ln(2) = -0.386...

6.  You can see that there isn't a turning point maximum in the range from x = 0 to x = 2, but the maximum value of the function in this range clearly happens when x = 0.  The value of the function here is f(0) = 0 - 2ln(0+1) = -2ln(1) = 0.

Hope this helps a bit more. 

29 abr 2014