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 4 - 2015 - Paper 3

For the graph of $y = |f(x)|$, we take the absolute value of the function:

1. Transformation: Since $f(x)$ has a minimum value of $-5$, $|f(x)|$ will reflect the portions of the graph below the x-axis above it.

2. Intercepts: The y-intercept remains at $(0, 3)$, whereas the x-intercept occurs at the same x-value as before:

• $x = \ln\left(\frac{5}{2}\right)$.

3. Asymptote: The horizontal asymptote at $y = -5$ becomes $y = 5$ in this case.

The final diagram reflects the changes, with the part of the graph below the x-axis flipped upwards.

Given that $f(x) = 2e^{x} - 5$, we can equate:

$2e^{x} - 5 = |f(y)|$

The values of $x$ that satisfy this equation depend on the nature of $f(y)$, either being positive or negative. Set conditions:

1. When $f(y) \geq 0$:
   $2e^{x} - 5 = 2e^{y} - 5 \Rightarrow e^{x} = e^{y} \Rightarrow x = y.$
2. When $f(y) < 0$:
   $2e^{x} - 5 = -(2e^{y} - 5) \Rightarrow 2e^{x} - 5 = -2e^{y} + 5.$
   Rearranging yields:
   $2e^{x} + 2e^{y} = 10 \Rightarrow e^{x} + e^{y} = 5.$

This indicates that for any given value $y$, we deduce sets of values for $x$ governed by the exponential decay.

To find the solutions where $|f(x)| = 2$, set up the equation:

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

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

The exact solutions are therefore:

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