# What is the instantaneous rate of change of #f(x)=(x-5)^2 # at #x=2#?

The instantaneous rate of change of

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To find the instantaneous rate of change of ( f(x) = (x - 5)^2 ) at ( x = 2 ), you need to compute the derivative of ( f(x) ) with respect to ( x ) and then evaluate it at ( x = 2 ).

The derivative of ( f(x) = (x - 5)^2 ) is ( f'(x) = 2(x - 5) ).

Now, plug in ( x = 2 ) into ( f'(x) ) to find the instantaneous rate of change:

[ f'(2) = 2(2 - 5) = 2(-3) = -6 ]

So, the instantaneous rate of change of ( f(x) = (x - 5)^2 ) at ( x = 2 ) is ( -6 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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