How to find instantaneous rate of change for #f(x)=x^3+x^2# at (2, 12)?
The instantaneous rate of change is given by the derivative.
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To find the instantaneous rate of change for ( f(x) = x^3 + x^2 ) at the point (2, 12), we can use the derivative of the function.
First, we find the derivative of ( f(x) ) with respect to ( x ), denoted as ( f'(x) ) or ( \frac{df}{dx} ).
( f'(x) = 3x^2 + 2x )
Then, we substitute ( x = 2 ) into the derivative to find the slope of the tangent line at the point (2, 12).
( f'(2) = 3(2)^2 + 2(2) = 12 + 4 = 16 )
So, the instantaneous rate of change of ( f(x) ) at ( x = 2 ) is 16.
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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.
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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