The derivative of $f(x) = 2x^3 + 6x^2 - 12x + 3$ can be expressed in the form $f^{ ext{'}}(x) = a(x + b)^2 + c$, where $a, b, c \in \mathbb{Z}$ and $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 5 - 2021

First, we differentiate $g(x) = 36x + 5$ to get:

$g^{ ext{'}}(x) = 36.$

Now, we set the inequality:

$f^{ ext{'}}(x) > g^{ ext{'}}(x) \Rightarrow 6x^2 + 12x - 12 > 36.$

Simplifying this:

$6x^2 + 12x - 48 > 0 \Rightarrow x^2 + 2x - 8 > 0.$

Factoring gives:

$(x+4)(x-2) > 0.$

Using a sign chart or test points, the solution for this inequality is:

$x < -4 \text{ or } x > 2.$

To find $k$, we first need to find the slope of the tangent line:

1. Differentiate $h(x)$: $h^{ ext{'}}(x) = 4 \cos(2x).$
2. Evaluate at $x = \frac{\pi}{6}$: $h^{ ext{'}}(\frac{\pi}{6}) = 4 \cos(\frac{\pi}{3}) = 4 \cdot \frac{1}{2} = 2.$

So, the slope of the tangent line at $A(\frac{\pi}{6}, h(\frac{\pi}{6}))$ is $2$.

3. The coordinates of the point $A(\frac{\pi}{6}, h(\frac{\pi}{6}))$: $h(\frac{\pi}{6}) = 2 \sin(\frac{\pi}{3}) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}.$ Hence, $A(\frac{\pi}{6}, \sqrt{3})$.

4. Using point-slope form of the line to find $k$: $y - \sqrt{3} = 2(x - \frac{\pi}{6}).$

5. Setting $x = 0$ to find where it intercepts the y-axis: $y - \sqrt{3} = 2(0 - \frac{\pi}{6}) \Rightarrow y = 2(-\frac{\pi}{6}) + \sqrt{3} = -\frac{\pi}{3} + \sqrt{3}.$

Calculating this gives: [ k \approx -0.68 \text{ (to two decimal places) }. ]