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Find the slope of the tangent line to the curve f(x)= x^2 + x - 1/x at x=-2
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You need to take the derivative of f(x) which represents you're curve. This will give you the slope as a function of x, and then you can plug in you're x value.
that is \[f'(x)=2x+1+\frac{1}{x^2}\] now plug in -2
you get \[f'(-2)=2\times -2+1+\frac{1}{(-2)^2}\]
i get \[-\frac{11}{4}\] as the slope
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