# What is the instantaneous rate of change of #f(x)=1/(x^3-2x+5 )# at #x=1 #?

This represents the derivative's value at x = 1.

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To find the instantaneous rate of change of the function ( f(x) = \frac{1}{x^3 - 2x + 5} ) at ( x = 1 ), we use the concept of the derivative. The derivative of a function at a specific point represents its instantaneous rate of change at that point.

To find the derivative of ( f(x) ), we'll use the quotient rule:

[ \frac{d}{dx}\left(\frac{1}{x^3 - 2x + 5}\right) = \frac{-3x^2 - 2}{(x^3 - 2x + 5)^2} ]

Now, to find the instantaneous rate of change at ( x = 1 ), we substitute ( x = 1 ) into the derivative expression:

[ \frac{-3(1)^2 - 2}{(1^3 - 2(1) + 5)^2} ]

Solving this expression gives us the instantaneous rate of change of ( f(x) ) at ( x = 1 ).

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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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