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MIT 18.01 Single Variable Calculus (OCW)
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OpenStudy (jkristia):
Hi, I have a problem understanding the solution to the problem in 1C-2. In the solution, how come f(a) becomes 0 ? 1C-2 Let f(z) = (z -a)g(z). Use the definition of the derivative to calculate that f'(a) = g(a).
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OpenStudy (jkristia):
I got the answer from another forum. f(a) becomes 0 because x -> a, meaning f(a) can be written as (x-a)g(a) = (a - a)g(a) = 0*g(a) = 0.
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