Least squares and integration problem

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Consider the vector space C[-1,1] with an inner product defined by
<f,g> = the integral from 1 to -1 of f(x)g(x) dx
a)
Show that
u1(x)= 1/(2^.5) u2(x)= ((6^.5)/2)x
form an orthonormal set of vectors
b)
Use the result from a) to find the best least squates approximation to
h(x)= x^(1/3) + x^(2/3)
by a linear function.
For part a) I have shown that u1 and u2 each have an inner product of zero and a length of one.
I've been trying to find a solution to b) in the form
p(x) = (c1)(u1(x)) + (c2)(u2(x))
where ci = the integral from -1 to 1 of (ui(x))h(x)
evaluating this integral for c1 produces
(1/(2^.5))[(3/4)x^(4/3) + (3/5)x^(5/3)] evaluated from -1 to 1
Now I've finally got to the problem
Putting -1 in would produce complex numbers and I don't know how to proceed.
I'm not sure weather the problem here is that I'm not aproaching the least squares problem correctly or weather I'm not approaching the integral correctly. Any help would be appreciated, thanks.
 
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Putting -1 in for x? That doesn't produce complex numbers. The cube root of -1 is -1. Its value at x= 1 is \frac{1}{\sqrt{2}}\left(\frac{3}{4}+\frac{3}{5}\right) and its value at x= -1 is \frac{1}{\sqrt{2}}\left(\frac{3}{4}-\frac{3}{5}\right) Subtracting gives just \frac{1}{\sqrt{2}}\frac{6}{5}.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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