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

Answer 1

The instantaneous rate of change of the function #= 7#

Given -

#y=x^2+3x-4#
The first derivative of the function gives the rate of change for any value of #x#
#dy/dx=2x+3#
At #x=2# the rate of change is -
At #x=2; dy/dx=2(2)+3=4+3=7#
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Answer 2

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

The derivative of the function ( x^2 + 3x - 4 ) with respect to ( x ) is ( 2x + 3 ).

Now, evaluate the derivative at ( x = 2 ): [ 2(2) + 3 = 4 + 3 = 7 ]

So, the instantaneous rate of change of the function ( x^2 + 3x - 4 ) when ( x = 2 ) is ( 7 ).

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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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