Question 38

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Jambmaths question: 

Find the minimal value of the function $y=x(1+x)$

Option A: 

$\tfrac{1}{4}$

Option B: 

$\tfrac{1}{2}$

Option C: 

$\tfrac{1}{4}$

Option D: 

$\tfrac{1}{2}$

Jamb Maths Solution: 

$\begin{align} & y=x(1+x) \\ & y=x+{{x}^{2}} \\ & \frac{dy}{dx}=1+2x \\ & \text{At stationary point }\frac{dy}{dx}=0 \\ & 1+2x=0;\text{   }x=-\frac{1}{2} \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=2>0,\text{ so at }x=-\frac{1}{2}\text{ is a min point} \\ & \text{The minimal value is} \\ & y=(-\tfrac{1}{2})+{{(\tfrac{1}{2})}^{2}}=-\tfrac{1}{2}+\tfrac{1}{4}=-\tfrac{1}{2} \\\end{align}$

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