# How do you find the instantaneous rate of change of the function #y=4x^3+2x-3# when x=2?

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To find the instantaneous rate of change of the function ( y = 4x^3 + 2x - 3 ) when ( x = 2 ), you can use the derivative of the function. The derivative of the function gives the slope of the tangent line to the function at any given point. So, you need to find the derivative of the function with respect to ( x ), and then evaluate it at ( x = 2 ).

First, find the derivative of the function:

[ \frac{dy}{dx} = 12x^2 + 2 ]

Next, evaluate the derivative at ( x = 2 ):

[ \frac{dy}{dx} \bigg|_{x=2} = 12(2)^2 + 2 ]

[ = 12(4) + 2 ]

[ = 48 + 2 ]

[ = 50 ]

So, the instantaneous rate of change of the function ( y = 4x^3 + 2x - 3 ) when ( x = 2 ) is ( 50 ).

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To find the instantaneous rate of change of the function y = 4x^3 + 2x - 3 when x = 2, we need to find the derivative of the function with respect to x, and then evaluate it at x = 2.

The derivative of the function y = 4x^3 + 2x - 3 with respect to x is given by:

dy/dx = 12x^2 + 2.

Now, we evaluate the derivative at x = 2:

dy/dx = 12(2)^2 + 2 = 12(4) + 2 = 48 + 2 = 50.

So, the instantaneous rate of change of the function y = 4x^3 + 2x - 3 when x = 2 is 50.

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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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