Sorry another question,
For f(x)=x^(p/q) where p is even and q is odd
Prove that when p>q there is a local min
And when p<q there is a cusp
Again thanks in advance
Assume p, q positive.
f'(x) = p/q * x^(p/q - 1) = p/q * x^((p-q)/q)
If p>q, then f'(0) = 0
Also, as q is odd, f(x) has an implied domain of all reals (you can take an odd root of a negative number)
Now, p is even, so f(-x) = f(x). It's an even function. Also note that f(x)>0 for all x not 0. Therefore, f(x) > 0 at either side of x=0, meaning you have a local minimum.
If p<q, f'(0) is not defined. In particular, as p is even and q is odd, p-q is odd. Thus, x^((p-q)/q) does not change the sign of x. Try it. (-1)^(3/5) is still negative. Similarly, f'(x)>0 for x>0
Therefore, the limit as x approaches zero from the left goes to -inf, but the limit from the right goes to +inf. The function is continuous at the origin though so we have a cusp.
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