The graph of the symmetric function $f(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} x^2}$ is shown below - Leaving Cert Mathematics - Question 8 - 2018

The area of the rectangle can be calculated using the formula:

$\text{Area} = \text{length} \times \text{width}$

Here, the length is the distance from A to B on the x-axis, which is 2 (from 0 to -1), and the width is the y-coordinate at B:

$\text{Width} = \frac{1}{\sqrt{2\pi}}$
Thus, the area is:

$\text{Area} = 2 \times \frac{1}{\sqrt{2\pi}} \approx 0.484.$

To show that $f'(x)$ is decreasing at point C, we first compute the derivative:

$f'(x) = \frac{d}{dx} \left( \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} x^2} \right) = -\frac{1}{\sqrt{2\pi}} x e^{-\frac{1}{2} x^2}.$
Next, evaluating at C where $x = 1$:

$f'(1) = -\frac{1}{\sqrt{2\pi}} (1)e^{-\frac{1}{2} (1)^2} < 0.$
Since $f'(x)$ is negative at C, it is indeed decreasing there.

To show a point of inflection at B, we need to examine the second derivative:

1. First, calculate the first derivative again: $f'(x) = -\frac{1}{\sqrt{2\pi}} x e^{-\frac{1}{2} x^2}.$
2. Now, we differentiate once more to find the second derivative: $f''(x) = -\frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} x^2} (1 - x^2).$
   Setting $f''(x) = 0$ gives: $1 - x^2 = 0\implies x = \pm 1.$
   Evaluating $f''(-1)$: $f''(-1) = 0$; hence, it changes sign at this x-coordinate, showing that $( -1, \frac{1}{\sqrt{2\pi}} )$ is indeed a point of inflection.