Question 27

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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