# How do you find the instantaneous rate of change for #f(x)= x^3 -2x# for [0,4]?

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To find the instantaneous rate of change for ( f(x) = x^3 - 2x ) on the interval [0,4], you can use calculus. The instantaneous rate of change is given by the derivative of the function ( f(x) ), which is the slope of the tangent line to the curve at a specific point.

First, find the derivative of ( f(x) ) with respect to ( x ) by using the power rule for differentiation:

[ f'(x) = 3x^2 - 2 ]

Now, to find the instantaneous rate of change at a specific point within the interval [0,4], you can plug the x-value into the derivative equation. For example, to find the instantaneous rate of change at ( x = 0 ):

[ f'(0) = 3(0)^2 - 2 = -2 ]

Similarly, to find the instantaneous rate of change at ( x = 4 ):

[ f'(4) = 3(4)^2 - 2 = 46 ]

So, the instantaneous rate of change at ( x = 0 ) is -2, and at ( x = 4 ) is 46.

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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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