Let $h(x) = ext{cos}(2x)$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 3 - 2018

Next, evaluate $h'\left( \frac{-\pi}{3} \right)$: $h'\left( \frac{-\pi}{3} \right) = -2 \sin\left( 2 \cdot \frac{-\pi}{3} \right) = -2 \sin\left( -\frac{2\pi}{3} \right) = -2 \left( -\frac{\sqrt{3}}{2} \right) = \sqrt{3}$

The slope of the tangent is given by the derivative, which we found to be $\sqrt{3}$. The angle $\theta$ can be found using the tangent function: $\tan(\theta) = \sqrt{3}$ Thus, the angle: $\theta = 60^\circ$

To find the average value of $h(x)$ over the interval $0 \leq x \leq \frac{\pi}{4}$, we use the formula: $\text{Average} = \frac{1}{b-a} \int_a^b h(x) \, dx$ Here, $a = 0$ and $b = \frac{\pi}{4}$: $\text{Average} = \frac{1}{\frac{\pi}{4} - 0} \int_0^{\frac{\pi}{4}} \cos(2x) \, dx$

The integral of $\cos(2x)$ is: $\int \cos(2x) \, dx = \frac{\sin(2x)}{2}$ Thus, $\int_0^{\frac{\pi}{4}} \cos(2x) \, dx = \left[ \frac{\sin(2x)}{2} \right]_0^{\frac{\pi}{4}} = \frac{\sin(\frac{\pi}{2})}{2} - \frac{\sin(0)}{2} = \frac{1}{2}$

Substituting back, we get: $\text{Average} = \frac{1}{\frac{\pi}{4}} \cdot \frac{1}{2} = \frac{4}{\pi} \cdot \frac{1}{2} = \frac{2}{\pi}$