An apache helicopter of enemy is flying along the curve given by  $y=x^2+7$. A Soldier, placed at (3, 7) wants to shoot down the helicopter when it is nearest to him. Then the nearest distance is …… (in units) 

For each value of ‘x’, the helicopter’s position is at a point  $\left(x,x^2+7\right)$. 

Therefore, the distance between the helicopter and soldier placed at 

$\left(3,7\right)is\sqrt{{\left(x-3\right)}^2+{\left(x^2+7-7\right)}^2},$ i.e., $\sqrt{{\left(x-3\right)}^2+x^4}$

$Letf\left(x\right)={\left(x-3\right)}^2+x^{4}orf\text{'}\left(x\right)=2\left(x-3\right)+4x^3=2\left(x-1\right)\left(2x^2+2x+3\right)$

Thus $f\text{'}\left(x\right)=0givesx=1or2x^2+2x+3=0$  for which there are no real roots. Also there are no end points of the interval to be added to the set for which $f\text{'}\left(x\right)=0$ i.e. there is only one end point, namely x = 1 at which $f\text{'}\left(x\right)=0.$ The value of ‘f’ at this point is given by $f\left(1\right)={\left(1-3\right)}^2+{\left(1\right)}^2=5.$ $\sqrt{f\left(0\right)}=\sqrt{{\left(0-3\right)}^2+{\left(0\right)}^2}=3>\sqrt{5},$It follows that $\sqrt{5}$ is the minimum value of $\sqrt{f\left(x\right)}$. Hence,  $\sqrt{5}$  is the minimum distance between the soldier and helicopter.