Let f be a real-valued differentiable function on R (the set of all real numbers) such that f(1)=1 . If the y -intercept of the tangent at any point P(x, y) on the curve y=f(x) is equal to the cube of the abscissa of P , then the value of f(-3) is equal to

Step-by-step solution

The equation of the tangent at (x, y) to the given curve y=f(x) is gathered Y-y= d y d x (X-x) Y -intercept =y-x d y d x gathered According to the question x^3=y-x d y d x d y d x - y x =-x^2 which is linear in x . IF =e^ -1 x d x = 1 x Required solution is aligned & y 1 x = -x^2 1 x d x & y x = -x^2 2 +c & y= -x^3 2 +c x & At x=1, y=1 , & 1= -1 2 +c & c= 3 2 & aligned Now, aligned f(-3) & = 27 2 + 3 2 (-3) & = 27-9 2 =9 aligned