Let & > 0, f: (-e, e) → R be a smooth function such that f'(0) = = 0. Let C be the curve given as the graph of f. A standard parametrization of Cis a(t) = (t, f(t)), te (-ɛ,ɛ). (a) Compute the curvature of C at t = 0, in terms of derivatives of f. (b) (*, optional) Suppose that f,g: (-e, e) are smooth functions such that: f(0) = g(0), f'(0) = g'(0) = 0, f"(x) ≥g"(x) > 0 for all x € (-e, e). Additionally, suppose that f(x) ≤ g(x) for all x € (-e, e). Show that f(x) = g(x) in (-ɛ, ɛ). Can you translate the conclusions of this statement in geometric language (i.e. in terms of positions of two curves and their curvature)?

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