Question 38

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Jambmaths question: 

Find the value of x at the minimum point of the curve $y={{x}^{3}}+{{x}^{2}}-x+1$

Option A: 

1

Option B: 

-1

Option C: 

$-\frac{1}{3}$

Option D: 

$\frac{1}{3}$

Jamb Maths Solution: 

$\begin{align}  & y={{x}^{3}}+{{x}^{2}}-x+1 \\ & \frac{dy}{dx}=3{{x}^{2}}+2x-1 \\ & \text{At stationary point }\frac{dy}{dx}=0 \\ & \therefore 3{{x}^{2}}+2x-1=0 \\ & (3x-1)(x+1)=0 \\ & x=-1\text{ or }x=\frac{1}{3} \\ & \text{To determine the minimum and maximum point} \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dx}(3{{x}^{2}}+2x-1)=6x+2 \\ & \text{At }x=\frac{1}{3} \\ & {{\left. \frac{{{d}^{2}}y}{d{{x}^{2}}} \right|}_{x=\tfrac{1}{3}}}=6(\tfrac{1}{3})+2=4\text{   (minimum point)} \\ & \text{At }x=1 \\ & {{\left. \frac{{{d}^{2}}y}{d{{x}^{2}}} \right|}_{x=1}}=6(-1)+2=-4\text{  (maximum point)} \\ & \text{The value of}x\text{at minimum point is}~\text{ }\frac{1}{3} \\\end{align}$

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