# What is the instantaneous rate of change of #f(x)=ln(2x^2-x+6) # at #x=-2 #?

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To find the instantaneous rate of change of ( f(x) = \ln(2x^2 - x + 6) ) at ( x = -2 ), you need to find the derivative of the function and then evaluate it at ( x = -2 ).

First, find the derivative of ( f(x) ): [ f'(x) = \frac{d}{dx} \ln(2x^2 - x + 6) ]

Apply the chain rule: [ f'(x) = \frac{1}{2x^2 - x + 6} \cdot \frac{d}{dx} (2x^2 - x + 6) ]

[ f'(x) = \frac{1}{2x^2 - x + 6} \cdot (4x - 1) ]

Now, evaluate ( f'(-2) ): [ f'(-2) = \frac{1}{2(-2)^2 - (-2) + 6} \cdot (4(-2) - 1) ]

[ f'(-2) = \frac{1}{8 + 2 + 6} \cdot (-8 - 1) ]

[ f'(-2) = \frac{1}{16} \cdot (-9) ]

[ f'(-2) = -\frac{9}{16} ]

So, the instantaneous rate of change of ( f(x) ) at ( x = -2 ) is ( -\frac{9}{16} ).

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