The time, in days of practice, it takes Jack to learn to type $x$ words per minute (wpm) can be modelled by the function: $$t(x) = k \left[ \ln \left( 1 - \frac{x}{80} \right) \right], \text{ where } 0 \leq x \leq 70, \ x \in \mathbb{R}, \text{ and } k \text{ is a constant.}$$ (a) Based on the function $t(x)$, Jack can learn to type 35 wpm in 35-96 days - Leaving Cert Mathematics - Question 7 - 2018

Using the derived formula, substitute 100 for $t(x)$:

$100 = -62.5 \left[ \ln \left( 1 - \frac{x}{80} \right) \right]$

Rearranging gives:

$-\frac{100}{-62.5} = \ln \left( 1 - \frac{x}{80} \right)$

Calculate:

$1.6 = \ln \left( 1 - \frac{x}{80} \right)$

Exponentiating both sides:

$1 - \frac{x}{80} = e^{-1.6}$

Thus,

$\frac{x}{80} = 1 - e^{-1.6}$

So,

$x = 80 \left( 1 - e^{-1.6} \right) \approx 64 \text{ wpm (to the nearest whole number)}$.

The function $p(x)$ is linear, passing through the origin. For values:

• When $x=0$, $p(0)=0$
• When $x=70$, $p(70)=1.5 \times 70 = 105$

Plot these points to draw the line.

The intersection point of $h(x)$ occurs where the graphs of $p(x)$ and $t(x)$ meet. Estimating from the graph gives approximately $x \approx 62$ wpm.

To find the maximum value:

Start with:

$h(x) = 1.5x - t(x)$

Taking the derivative:

$h'(x) = 1.5 - t'(x)$

Setting $h'(x) = 0$:

Identify $x$ value yielding the maximum (calculating $t'(x)$ needed).

At $x \approx 38.3$, find:

$h(38.3) = 1.5(38.3) - t(38.3) \approx 17$

Thus, the maximum value of $h(x)$ is $17$ days.