How do you find the instantaneous rate of change for #f(x)= (x^2-2)/(x-1)# for x=2?

Answer 1

Differentiate using the quotient rule and then use the value of x to get #-2#

The rate of change is the derivative of the function. To find this we use the quotient rule #d/dx(f(x)/g(x)) =(g(x)f'(x) - g'(x)f(x))/(g(x)^2)#
#d/dx(f(x)/g(x)) = ((x-1)*2x - 1*(x^2 -2))/(x-1)^2#
#d/dx(f(x)/g(x)) = (2x^2 - 2x -x^2 +2)/(x-1)^2 = (x^2 - 2x - 2)/(x-1)^2#
If# x=2# the rate of change # (2^2 -2*2-2)/(2-1)^2 =-2/1 = -2#
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Answer 2

To find the instantaneous rate of change for ( f(x) = \frac{x^2 - 2}{x - 1} ) at ( x = 2 ), you can use the derivative of the function.

First, find the derivative of ( f(x) ) using the quotient rule, then evaluate the derivative at ( x = 2 ) to get the instantaneous rate of change at that point.

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Answer from HIX Tutor

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.

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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