An aerial view of a stretch of road is shown in the diagram below - NSC Mathematics - Question 9 - 2017 - Paper 1

Differentiate:

$\frac{d(PB)}{dx} = \frac{1}{2PB}\cdot 2(x + (x^2 - 1)(2x))$

Set the derivative to zero to find critical points:

$2x(x^2 - 1) + (x^2 - 1) \cdot 2x = 0$

Factoring gives:

$x(2x^2 - 1) = 0$

This implies either $x = 0$ or $x = \frac{1}{\sqrt{2}}$. We discard $x = 0$ since it does not yield the closest point.

Now, substitute $x = \frac{1}{\sqrt{2}}$ back into the distance function:

$PB = \sqrt{\left(\frac{1}{\sqrt{2}} - 0\right)^2 + \left(\left(\frac{1}{\sqrt{2}}\right)^2 + 2 - 3\right)^2}$

Calculating gives:

$PB = \sqrt{\frac{1}{2} + (0.5 - 1)^2} = \sqrt{\frac{1}{2} + \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} = 0.87$

Thus, the distance between Benny and the car, when the car is closest, is approximately 0.87 units.