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Let f be a real-valued differentiable...

Let `f` be a real-valued differentiable function on `R` (the set of all real numbers) such that `f(1)=1.` If the `y-in t e r c e p t` 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________

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Equation of tangent to `y=f(x)` at point `(x_(1),y_(1))` is
`y-y_(1)=m(x-x_(1))`
Put `x=0` to get y intercept.
`y_(1)-mx_(1)=x_(1)^(3)` (given)
`therefore y_(1)-x_(1)(dy)/(dx)=x_(1)^(3)`
`x(dy)/(dx) -y=-x^(3)`
`therefore (dy)/(dx) -y/x =-x^(2)`
I.F. `=e^(int-1/xdx) = e^-("ln "x)=1/x`
Thus, solution is `y xx 1/x = int-x^(2) xx 1/x dx`
or `y/x =-x^(2)/2 + c`
or `f(x) = -x^(2)/2+3/2x` [as `f(1)=1]`
`therefore f(-3)=9`
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