(a) (i) $f(x) = \frac{2}{e^x}$ and $g(x) = e^x - 1$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 3 - 2016

Using the values obtained in part (i), plot the points for each function on the provided grid:

• For $f(x)$, plot the points:

  • $(0, 2)$
  • $(0.5, 1.21)$
  • $(1, 0.74)$
  • $(\ln(4), 0.5)$

• For $g(x)$, plot the points:

  • $(0, 0)$
  • $(0.5, 0.65)$
  • $(1, 1.72)$
  • $(\ln(4), 3)$

Make sure to label the graphs clearly, indicating which line corresponds to $f(x)$ and which corresponds to $g(x)$.

From the graphs plotted in part (ii), visually inspect where the two curves intersect. This will give an approximate value of $x$. After checking the graphs, it appears that the value is around $x \approx 0.7$. Make a note to specify this estimated value clearly, even if it is not exactly stated on the graph.

To solve the equation $f(x) = g(x)$:

1. Substitute the expressions for $f(x)$ and $g(x)$:
   $\frac{2}{e^x} = e^x - 1$

2. Multiply both sides by $e^x$ (valid for $e^x \neq 0$):
   $2 = e^{2x} - e^x$

3. Rearrange as:
   $e^{2x} - e^x - 2 = 0$
   Let $y = e^x$. This gives:
   $y^2 - y - 2 = 0$

4. Factor the quadratic:
   $(y - 2)(y + 1) = 0$
   Hence, $y = 2$ or $y = -1$ (not valid as $e^x > 0$).

5. Therefore, take $y = e^x = 2$:
   $x = \ln(2) \approx 0.693$
   Hence, the solution is $x = \ln(2)$.