Question

A curve has equation y=8/x +2x.
(i) Find the dy/dx and d^2y/dx^2.
(ii) Find the coordinates of the stationary points and state,with a reason,the nature of each stationary point.

Answer 1

Given,

$\longrightarrow\sf{y=\dfrac{8}{x}+2x}$

Then,

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{d}{dx}\left[\dfrac{8}{x}+2x\right]}$

$\longrightarrow\sf{\dfrac{dy}{dx}=8\cdot\dfrac{d}{dx}\left(\dfrac{1}{x}\right)+2\cdot\dfrac{dx}{dx}}$

$\longrightarrow\sf{\dfrac{dy}{dx}=8\cdot\dfrac{d}{dx}\left(x^{-1}\right)+2\cdot1}$

$\longrightarrow\sf{\dfrac{dy}{dx}=8(-x^{-2})+2}$

$\longrightarrow\sf{\underline{\underline{\dfrac{dy}{dx}=-\dfrac{8}{x^2}+2}}}$

And,

$\longrightarrow\sf{\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left[-\dfrac{8}{x^2}+2\right]}$

$\longrightarrow\sf{\dfrac{d^2y}{dx^2}=-8\cdot\dfrac{d}{dx}\left(\dfrac{1}{x^2}\right)}$

$\longrightarrow\sf{\dfrac{d^2y}{dx^2}=-8\cdot\dfrac{d}{dx}\left(x^{-2}\right)}$

$\longrightarrow\sf{\dfrac{d^2y}{dx^2}=-8\cdot-2\left(x^{-3}\right)}$

$\longrightarrow\sf{\underline{\underline{\dfrac{d^2y}{dx^2}=\dfrac{16}{x^{3}}}}}$

The stationary points of $\sf{y}$ are the points where $\sf{\dfrac{dy}{dx}=0.}$ Then,

$\longrightarrow\sf{\dfrac{dy}{dx}=0}$

$\longrightarrow\sf{-\dfrac{8}{x^2}+2=0}$

$\longrightarrow\sf{\dfrac{8}{x^2}=2}$

$\longrightarrow\sf{x^2=4}$

$\longrightarrow\sf{x=\pm2}$

For $\sf{x=2,}$

$\longrightarrow\sf{y=\dfrac{8}{2}+2\cdot2}$

$\longrightarrow\sf{y=8}$

For $\sf{x=-2,}$

$\longrightarrow\sf{y=\dfrac{8}{-2}+2\cdot-2}$

$\longrightarrow\sf{y=-8}$

So the points $\sf{(-2,\ -8)}$ and $\sf{(2,\ 8)}$ are the stationary points.