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(C) The equation of the tangent at point $P(x, y)$ is $Y - y = \frac{dy}{dx}(X - x)$.To find the $y$-intercept,set $X = 0$,which gives $Y = y - x \fra

Vedclass · Accepted Answer

$9$

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(C) The equation of the tangent at point $P(x, y)$ is $Y - y = \frac{dy}{dx}(X - x)$.To find the $y$-intercept,set $X = 0$,which gives $Y = y - x \frac{dy}{dx}$.According to the problem,the $y$-intercept is equal to the cube of the abscissa,so $y - x \frac{dy}{dx} = x^3$.Rearranging this gives $x \frac{dy}{dx} - y = -x^3$,or $\frac{dy}{dx} - \frac{y}{x} = -x^2$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{1}{x}$ and $Q(x) = -x^2$.The integrating factor is $IF = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$.The solution is $y \cdot \frac{1}{x} = \int (-x^2) \cdot \frac{1}{x} dx = \int -x dx = -\frac{x^2}{2} + C$.Thus,$f(x) = -\frac{x^3}{2} + Cx$.Given $f(1) = 1$,we have $1 = -\frac{1}{2} + C$,which implies $C = \frac{3}{2}$.So,$f(x) = -\frac{x^3}{2} + \frac{3}{2}x$.Calculating $f(-3) = -\frac{(-3)^3}{2} + \frac{3}{2}(-3) = -\frac{-27}{2} - \frac{9}{2} = \frac{18}{2} = 9$.

Let $f$ be a real-valued differentiable function on $\mathbb{R}$ (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

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