InterviewSolution
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InterviewSolution
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Find the co-ordinates of the point on curve `y=(x^2-1)/(x^2+1), (x>0)` where the gradient of the tangent to the curve is maximum. |
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Answer» Here, equaton of the curve, `y = (x^2-1)/(x^2+1)` `:. dy/dx = ((x^2+1)(2x)-(x^2-1)(2x))/(x^2+1)^2` `=>dy/dx = (4x)/(x^2+1)^2` We have to find a point, where slope is maximum. `:. (d^2y)/dx^2` should be `0`. `=>(d^2y)/dx^2 = ((x^2+1)^2 4-4x(2(x^2+1))(2x) )/(x^2+1)^4 ` `=>((x^2+1)^2 4-4x(2(x^2+1))(2x) )/(x^2+1)^4 =0` `=>(x^2+1)^2(4-16x^2) = 0` As `x^2+1` can not be `0`, `:. 4 - 16x^2 = 0` `=> x = +-1/2` We will take `x = 1/2` as `x gt 0`. `:. y = (x^2-1)/(x^2+1) = (1/4-1)/(1/4+1) = 3/5` So, the point at which slope is maximum is `(1/2,3/5)`. |
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