Question

If the function $y=f(x)$ is represented as,
$x=\varphi (t)=t^3-5t^2-20t+7$
$y=\psi (t)=4t^3-3t^2-18t+3(|t|<2),$ then:

• A. $y_{\max }=12$
• B. $y_{\max }=14$
• C. $y_{\min }=-67/4$
• D. $y_{\min }=-69/4$

Answer 1

Answer: (B), (D)

we have
$x=t^5-5t^3-20t+7$
and, $y=4t^3-3t^2-18t+3$
$\therefore \frac{dx}{dt}=5t^4-15t^2-20$
and $\frac{dy}{dt}=12t^2-6t-18$
$\Rightarrow \frac{dx}{dt}=5\left(t^2-4\right)\left(t^2+1\right)$ and
$\Rightarrow \frac{dy}{dt}=6(2t-3)(t+1)$
$\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{6(2t-3)(t+1)}{5\left(t^2-4\right)\left(t^2+1\right)}$
Now, $\frac{dy}{dx}=0$
$\Rightarrow t=\frac{3}{2},-1$
clearly, these values satisfy $-2<t<2$.
Now, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right)$
$=\frac{d}{dt}\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right)\cdot \frac{dt}{dx}$
$=\frac{\frac{dx}{dt}\cdot \frac{d^{2}y}{d{t}^2}-\frac{dy}{dt}\cdot \frac{d^{2}x}{d{t}^2}}{{\left(\frac{dx}{dt}\right)}^3}$
$=\frac{\left(5t^4-15t^2-20\right)(24t-6)-\left(12t^2-6t-18\right)\left(20t^3-30t\right)}{{\left(5t^4-15t^2-20\right)}^3}$
When $t=-1$, we have
$\frac{d^{2}y}{d{x}^2}=\frac{-30\times -30}{(-30{)}^3}<0,x=31$ and $y=14$
when $t=\frac{3}{2}$, we have
$\frac{d^{2}y}{d{x}^2}>0,x=\frac{-1033}{32}$ and $y=\frac{-69}{4}$
Hence, the function $y=f(x)$ attains a local maximum at $(31,14)$ and a local minimum at $(-1033/32,-69/4)$