The slope of the tangent to a curve y = f(x) at any point (x, y) is given by 2(y+2) x x , where x > 1 . If the curve passes through the point (e, 2) , then the value of f(e^3) is equal to

Step-by-step solution

The given condition can be written as a differential equation: dy dx = 2(y+2) x x Separating the variables, we get: dy y+2 = 2 x x dx Integrating both sides: dy y+2 = 2 1 x x dx (y+2) = 2 ( x) + C It is given that the curve passes through (e, 2) . Substituting x = e and y = 2 : (2+2) = 2 ( e) + C 4 = 2 (1) + C C = 4 Substituting the value of C back into the equation: (y+2) = 2 ( x) + 4 (y+2) = (( x)^2) + 4 (y+2) = (4( x)^2) Therefore, y + 2 = 4( x)^2 f(x) = 4( x)^2 - 2 To find f(e^3) , substitute x = e^3 : f(e^3) =