Given that $$f(x) = 2e^{x} - 5, \, x \in \mathbb{R}$$ (a) sketch, on separate diagrams, the curve with equation (i) $y = f(x)$ (ii) $y = |f(x)|$ On each diagram, show the coordinates of each point at which the curve meets or cuts the axes - Edexcel - A-Level Maths Pure - Question 2 - 2015 - Paper 3

The function $y = |f(x)|$ reflects any negative values of $f(x)$ above the x-axis. Since $f(x)$ is negative between $x = -\infty$ and $x = \ln\left(\frac{5}{2}\right)$, we'll need to find where the curve intersects the x-axis and reflect them.

1. Identify the positive and negative regions:

   • For $x < \ln\left(\frac{5}{2}\right)$, $f(x) < 0$; hence $|f(x)| = -f(x) = 5 - 2e^{x}$.
   • For $x \geq \ln\left(\frac{5}{2}\right)$, $f(x) \geq 0$; hence $|f(x)| = f(x) = 2e^{x} - 5$.

2. Find x-intercepts for $|f(x)|$:
   As previously found, the intercept happens at:

   • $x = \ln\left(\frac{5}{2}\right)$ at $(\ln\left(\frac{5}{2}\right), 0)$.

3. Find the y-intercept as before: $f(0) = -3$ which reflects to $(0, 3)$ in $|f(x)|$.

4. Asymptote: The horizontal asymptote remains $y = 5$ as $x \to \infty$.

The curve for $y = |f(x)|$ will touch the x-axis at $(\ln\left(\frac{5}{2}\right), 0)$, cross through $(0, 3)$, and approach $y = 5$.

Given the equation:
$f(x) = |f(y)|$

1. Analyze $f(x)$ and recognize:

   • If $f(x) \geq 0$ then $|f(y)| = f(y)$.
   • If $f(x) < 0$ then $|f(y)| = -f(y)$.

2. Solve for these cases.

   • For $f(x) \geq 0$:
     $2e^{x} - 5 = 2e^{y} - 5 \Rightarrow e^{x} = e^{y} \Rightarrow x = y$
   • For $f(x) < 0$: $2e^{x}-5 = - (2e^{y} - 5) \Rightarrow 2e^{x} = 2(2e^{y}) \Rightarrow e^{x} = 2 e^{y} \Rightarrow x = y + \ln(2)$

3. Thus, the set of values will consist of points where $x = y$ and also locations where $x = y + \ln(2)$ where $f(x) \geq 0$.

To find the exact solutions for the equation:
$|f(x)| = 2$

1. Set up the two scenarios:

   • When $f(x) = 2$: $2e^{x} - 5 = 2 \Rightarrow 2e^{x} = 7 \Rightarrow e^{x} = \frac{7}{2} \Rightarrow x = \ln\left(\frac{7}{2}\right)$

   • When $f(x) = -2$: $-(2e^{x} - 5) = 2 \Rightarrow -2e^{x} + 5 = 2 \Rightarrow -2e^{x} = -3 \Rightarrow e^{x} = \frac{3}{2} \Rightarrow x = \ln\left(\frac{3}{2}\right)$

2. The exact solutions are:

   • $x = \ln\left(\frac{7}{2}\right)$
   • $x = \ln\left(\frac{3}{2}\right)$