Subscribe Us

Header Ads

Differentation (Application/ Nature of stationary point)

\CIE\0606\2020\w\paper 12 \no 9

A curve has equation $y=(2 x-1) \sqrt{4 x+3}$.

(a) Show that $\dfrac{\mathrm{d} y}{\mathrm{~d} x}=\dfrac{4(A x+B)}{\sqrt{4 x+3}}$, where $A$ and $B$ are constants.

(b) Hence write down the $x$ -coordinate of the stationary point of the curve.

(c) Determine the nature of this stationary point.


%%%%%%%%% Solution %%%%%%%%% 


(a) $\begin{array}[t]{rcll} \dfrac{dy}{dx}&=&(2x-1) \dfrac{d}{dx} (4x+3)^{\dfrac{1}{2}} +(4x+3)^{\dfrac{1}{2}} \dfrac{d}{dx}(2x-1)\\ &=&(2x-1)\dfrac{1}{2} (4x+3)^{-\dfrac{1}{2}} \times 4 +\funv \times 2\\  &=&\dfrac{2(2x-1)}{\funvo}+2\funvo\\ &=& \dfrac{(4x-2)+(8x+6)}{\funvo}\\ &=& \dfrac{12x+4}{\funvo}\\ &=&\dfrac{4(3x+1)}{\funvo}\end{array}$

(b) For stationary point, $\dfrac{dy}{dx} =0$. Thus $\dfrac{4(3x+1)}{\funvo}=0\Rightarrow x=-\dfrac{1}{3}$.

(c) $\begin{array}[t]{|c||c|c|c|}\hline x& <-\dfrac{1}{3} & =-\dfrac{1}{3}&>-\dfrac{1}{3}\\ \hline 4(3x+1)&-&0&+\\ \funvo& +&+&+\\ \hline \dfrac{dy}{dx}&-&0&+\\ \hline\end{array}$ 
Thus the stationary point is minimum.

Post a Comment