It's clear that because of symmetry, the derivatives at -1/2, -1 and -2
will be the negative of those at 1/2, 1 and 2 for the function f(x) = x^2 since that is the graph of a parabola. Therefore I get f'(-1/2)
= -1, f'(-1) = -2 and f'(-2) = -4

Applied the definition to get the derivative of x^2. Put in (x + h) for x
in f(x) to get (x+ h)^ then subtract off f(x) which is x^2 and divide
the whole thing by h. Remember to expand (x+h)^2, which is (x+h)(x+h),
then simplify and put in 0 for h to get 2x, which matches the guess in
part c

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