Question 1163275
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Find the coordinates of the point of tangency:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(x,\,f(x)\)\|_{x=1}\ =\ \(1,\,(2(1)^2\,-\,1\)^{-5}\)\ =\ \(1,\,(1)^{-5}\)\ =\ (1,1)]


Let *[tex \Large u\ =\ \(2x^2\,-\,1\)], then *[tex \Large y\ =\ u^{-5}], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ \frac{dy}{du}\,\cdot\,\frac{du}{dx}\ =\ -6\(2x^2\,-\,1\)^{-6}\(4x\)]


Then 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\|_{x=1}\ =\ -6(1)^{-6}\(4(x)\)\ =\ -24]


So write a linear equation that represents a line that goes through the point (1,1) with a slope of -24.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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