The function $h(x)$ can be used to model the height above level ground (in metres) of a section of the path followed by a rollercoaster track, where $x$ is the horizontal distance from a fixed point - Leaving Cert Mathematics - Question b - 2021

To find the maximum height, we set the derivative $h'(x)$ equal to zero:

$0.003x^2 - 0.24x + 3.6 = 0$

Solving this quadratic equation will give the critical points. You can apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 0.003$, $b = -0.24$, and $c = 3.6$.

Calculating the discriminant:

$D = (-0.24)^2 - 4\cdot(0.003)(3.6)$

We find the roots of the equation, leading to:

$h'(20) = 0$

Verifying that this point is a local maximum can be done via the second derivative test or by establishing that $h'(x)$ changes from positive to negative at $x = 20$.

To find the inflection point, we need to determine where the second derivative $h''(x)$ changes sign:

First, we compute the second derivative:

$h''(x) = 0.006x - 0.24$

Setting $h''(x) = 0$ gives:

$0.006x - 0.24 = 0$

This simplifies to:

$x = 40$

We now evaluate $h(40)$ to find the height above the ground:

$h(40) = 0.001(40)^3 - 0.12(40)^2 + 3.6(40) + 5$

Upon calculating, we find:

$h(40) = 21 \, ext{metres}$