The derivative of 𝑓(𝑥) = 2𝑥³ + 6𝑥² − 12𝑥 + 3 can be expressed in the form 𝑓′(𝑥) = 𝑎(𝑥 + 𝑏)² + 𝑐, where 𝑎, 𝑏, 𝑐 ∈ ℤ and 𝑐 ∈ ℝ - Leaving Cert Mathematics - Question 5 - 2021

First, we find the derivative of $g(x)$:

$g'(x) = 36$

Now we set up the inequality:

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

Rearranging gives:

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

Factoring the quadratic:

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

The roots are $x = 2$ and $x = -4$. The inequality is satisfied for:

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

To find the value of $k$, we first differentiate $h(x)$:

$h'(x) = 4 \cos(2x)$

Now we evaluate at $x = \frac{\pi}{6}$:

$h'\left(\frac{\pi}{6}\right) = 4 \cos\left(\frac{\pi}{3}\right) = 4 \cdot \frac{1}{2} = 2$

Next, we find $h(\frac{\pi}{6})$:

$h\left(\frac{\pi}{6}\right) = 2 \sin\left(\frac{\pi}{3}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$

Using the point-slope form for the tangent line, we have:

$y - h(\frac{\pi}{6}) = h'\left(\frac{\pi}{6}\right)(x - \frac{\pi}{6})$

Substituting $h(\frac{\pi}{6}) = \sqrt{3}$ and the slope:

$y - \sqrt{3} = 2\left( x - \frac{\pi}{6} \right)$

To find the y-intercept (where $x = 0$):

$y - \sqrt{3} = 2(-\frac{\pi}{6}) \Rightarrow y = 2(-\frac{\pi}{6}) + \sqrt{3}$

Calculating this gives us:

$k \approx 0.68$