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find tangent to witch of agnesi y= 8/(4+x^2) at point (2,1)
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did you find the derivative?
\[\frac{d}{dx}\frac{1}{f(x)}=\frac{-f'(x)}{f^2(x)}\] is a start
would the derivative be y= (-16x)/(x^2+4)^2
yes, looks good
how to i proceed?
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find the slope by taking \(f'(2)\) then use the point slope formula for the equation of the line
so would it be y=1/2x
i don't know i didn't do it, but i can check
slope should be negative though
i get a slope of \(-\frac{1}{2}\) and the line as \(y=-\frac{1}{2}x+2\)
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how do i make the tangent of that line at (2,1)???
i used \(y-y_1=m(x-x_1)\) with \(m=-\frac{1}{2},x_1=2,y_1=1\)
when i graphed it though, the tangent was at a point 2, 7 or something strange like that
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