If y = f(x) is a monotonically increasing fun | vedclass

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(D) Let $u = \frac{dy}{dx}$. The given equation is $u^2 = 6 - u$, which implies $u^2 + u - 6 = 0$.Factoring the quadratic equation, we get $(u + 3)(u

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(D) Let $u = \frac{dy}{dx}$. The given equation is $u^2 = 6 - u$, which implies $u^2 + u - 6 = 0$.Factoring the quadratic equation, we get $(u + 3)(u - 2) = 0$.Thus, $u = -3$ or $u = 2$.Since $y = f(x)$ is a monotonically increasing function, $\frac{dy}{dx} > 0$, so we must have $\frac{dy}{dx} = 2$.Integrating $\frac{dy}{dx} = 2$ with respect to $x$, we get $y = 2x + C$.Using the initial condition $y(0) = 5$, we have $5 = 2(0) + C$, which gives $C = 5$.Therefore, the function is $y = 2x + 5$.To find $y(3)$, substitute $x = 3$: $y(3) = 2(3) + 5 = 6 + 5 = 11$.

If $y = f(x)$ is a monotonically increasing function such that $(\frac{dy}{dx})^2 = 6 - \frac{dy}{dx}$ and $y(0) = 5$, then $y(3) = \dots$

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