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

The area of the shaded rectangle can be found using the formula for area:

$ext{Area} = ext{width} imes ext{height}$

In this case, the width is 2 (distance from $-1$ to $1$) and the height is -rac{1}{ ext{sqrt}(2 ext{pi})}. Therefore:

ext{Area} = 2 imes ig(-rac{1}{ ext{sqrt}(2 ext{pi})}ig) = -rac{2}{ ext{sqrt}(2 ext{pi})}

To correct it to positive value for area:

Area = $0.484$ units² (to 3 decimal places).

To determine if the function is decreasing at point C, we first find the derivative of $f(x)$:

f'(x) = rac{1}{ ext{sqrt}(2 ext{pi})} e^{-rac{1}{2}x^2} (-x)

At $x = 1$ (point C):

f'(1) = rac{1}{ ext{sqrt}(2 ext{pi})} e^{-rac{1}{2}(1)^2} (-1)

Since $f'(1)$ is negative:

$f'(1) < 0$

This indicates that $f(x)$ is decreasing at point C.

To find the point of inflection, we need the second derivative:

f''(x) = rac{1}{ ext{sqrt}(2 ext{pi})} e^{-rac{1}{2}(-x)}

Calculating $f''(-1)$:

f''(-1) = rac{1}{ ext{sqrt}(2 ext{pi})} e^{-rac{1}{2}(-1)^2} > 0

Indicating that the concavity changes, thus confirming a point of inflection at B.