Jambmaths question:
Find the turning point of the function $y={{x}^{3}}-{{x}^{2}}-x$
Option A:
$1,-\tfrac{1}{3}$
Option B:
$-1,-\tfrac{1}{3}$
Option C:
$-1,\tfrac{1}{3}$
Option D:
$1,\tfrac{1}{3}$
Jamb Maths Solution:
$\begin{align} & y={{x}^{3}}-{{x}^{2}}-x \\ & \frac{dy}{dx}=3{{x}^{2}}-2x-1 \\ & \text{At turning point }\frac{dy}{dx}=0 \\ & 3{{x}^{2}}-2x-1=0 \\ & 3{{x}^{2}}-3x+x-1=0 \\ & 3x(x-1)+1(x-1)=0 \\ & (x-1)(3x+1)=0 \\ & x=1,\text{ }x=-\tfrac{1}{3} \\\end{align}$
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