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:

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

Where:

• Length = 2 (as it spans from -1 to 1 on the x-axis)
• Width = ( y )-coordinate at B = ( \frac{1}{2\sqrt{2\pi}} )

Calculating the area:

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

Converting to decimal form gives approximately 0.484. Thus, the area of the shaded rectangle is:

$0.484\text{ units}^2$

First, we find the first derivative of the function:

$f'(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} \cdot (-x)$

We are interested in the sign of $f'(x)$ at point C.

At $x = 1$:

$f'(1) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}(1)^2} (-1) < 0$

Therefore, since $f'(1) < 0$, we conclude that $f(x)$ is decreasing at point C.

To find the points of inflection, we calculate the second derivative:

$f''(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} [(-x)^{2} - 1]$

At $x = -1$:

f''(-1) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}(-1)^2} igg[(-1)^{2} - 1\bigg] = 0

Since $f''(-1) = 0$ and changes its sign around this value, we conclude that there is a point of inflection at B.