Question

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

Answer 1

`-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`