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Let f: R -> R and g: R -> R be differentiable, with g(x) ≠ 0...

Let f: R -> R and g: R -> R be differentiable, with g(x) ≠ 0 for all x. Assume that g(x) f'(x) = f(x) g'(x) for all x. Show that there is a real number c such that f(x) = cg(x) for all x. (Hint: Look at f/g.)

Let g: [0, ∞) -> R, with g(x) = x2 for all x ≥ 0. Let L be the line tangent to the graph of g that passes through the point (0, -6). Find the point (x0, x02) that is the point of tangency.

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