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Find the equations of all lines having slope 0 which are tangent to the curve `y=1/(x^2-2x+3)`. |
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Answer» Slope of tangent = 0 Equation of curve `y= 1/(x^(2)-2x+3)` ….(1) ` rArr (dy)/(dx) = (-1)/((x^(2)-2x+3)^(2)) d/(dx) (x^(2)-2x+3)` `=(-2(x-1))/((x^(2)-2x+3)^(2))` Slope of tangent at point `(x, y) = (-2(x-1))/((x^(2)-2x+3)^(2))` `:. (-2(x-1))/((x^(2)-2x+3)^(2))=0` ` rArr x = 1` put x = 1 in equation (1) ` y = 1/(1-2+3) = 1/2` equation of tangent at point ` (1,1/2)` ,brgt ` y - 1/2 = 0 (x-1)` ` rArr y = 1/2` |
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