How do you find the instantaneous rate of change of y=x^2-4x at x=1 by using the limit method?

1 Answer
Bill K.
Jul 27, 2015

-2

Explanation:

The instantaneous rate of change of y=f(x)=x^2-4x at x=1 is the derivative f'(1). Using the definition of the derivative in terms of limits, we have:

f'(1)=lim_{h->0}(f(1+h)-f(1))/h

=lim_{h->0}((1+h)^2-4(1+h)-(1-4))/h

=lim_{h->0}(1+2h+h^2-4-4h+3)/h

=lim_{h->0}(h(h-2))/h=lim_{h->0}(h-2)=0-2=-2

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