Question

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 _____

Answer 1

Answer: 9

$y-y_1=m\left(x-x_1\right)$
Put $x=0$, to get $y$ intercept
$y_1-m{x}_1=x_1^3$
$y_1-x_1\frac{dy}{dx}=x_1^3$
$x\frac{dy}{dx}-y=-x^3$
$\frac{dy}{dx}-\frac{y}{x}=-x^2$
$e^{\int \frac{1}{x}dx}=e^{-\ln x}=\frac{1}{x}$
$y\times \frac{1}{x}=\int -x^2\times \frac{1}{x}dx$
$\frac{y}{x}=-\int xdx$
$\Rightarrow \frac{y}{x}=-\frac{x^2}{2}+c$
$\Rightarrow f(x)=-\frac{x^3}{2}+\frac{3}{2}x$
$\therefore f(-3)=9$