f(x) = x^3 - 3x + 5. I know the derivative is 3x - 3 but I need to prove it using the definition above. When I do it and cancel out terms I end up with a negative 3 and an extra x in the derivative (which cannot happen!). Help!
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f(x) = x^3 - 3x + 5
f(x+h) = (x+h)^3 - 3(x+h) + 5 = x^3 + 3(x^2)h + 3x(h^2) + h^3 - 3x - 3h + 5
f(x+h) - f(x) = x^3 + 3(x^2)h + 3x(h^2) + h^3 - 3x - 3h + 5 - (x^3 - 3x + 5) = 3hx^2 + 3xh^2 + h^3 - 3h
[f(x+h) - f(x)] / h = 3x^2 + 3xh + h^2 - 3
lim h-> 0 [f(x+h) - f(x)] / h = 3x^2 + 3x(0) + (0)^2 - 3 = 3x^2 - 3
f(x+h) = (x+h)^3 - 3(x+h) + 5 = x^3 + 3(x^2)h + 3x(h^2) + h^3 - 3x - 3h + 5
f(x+h) - f(x) = x^3 + 3(x^2)h + 3x(h^2) + h^3 - 3x - 3h + 5 - (x^3 - 3x + 5) = 3hx^2 + 3xh^2 + h^3 - 3h
[f(x+h) - f(x)] / h = 3x^2 + 3xh + h^2 - 3
lim h-> 0 [f(x+h) - f(x)] / h = 3x^2 + 3x(0) + (0)^2 - 3 = 3x^2 - 3