Given that $f(x) = 2e^{x} - 5,\quad x \in \mathbb{R}$ (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 second curve to sketch involves taking the absolute value of the function we graphed previously.

1. Find the transformation:

   • The part of the graph where (f(x) < 0), specifically from negative infinity to the x-intercept, will reflect above the x-axis. The curve in this region will be represented as (y = -f(x)).

2. Determine key points:

   • The previous x-intercept becomes a minimum point: $(\ln{\frac{5}{2}}, 0)$.
   • The y-intercept remains at $(0, 3)$ (reflecting from $(0, -3)$).

3. Asymptote remains unchanged:

   • The horizontal asymptote is still (y = 5) as the function approaches this level as x tends to either direction.

To find the values where (f(x) = |f(y)|), we must analyze both cases where (f(y)) is positive and negative.

1. **When $f(y) \geq 0$:

   • Here, we can directly set $f(x) = f(y)$, thus leading to $x = y$.

2. **When $f(y) < 0$:

   • In this case, we reflect and equate:

   $f(x) = -f(y)$

   This leads again to a relationship involving both variables, but based on the structure of $f(y)$ (it approaches -5 and intercepts the x-axis at $(\ln\left(\frac{5}{2}\right), 0)$), we can derive that it holds across the intersection values accordingly.

   • Thus, deduced $x$ values will lie in the range $x \in ( -\infty, \ln{\frac{5}{2}} ) \cup (\ln{\frac{5}{2}}, +\infty)$.

To solve for (|f(x)| = 2):

1. Set up the equations based on the two cases from absolute values:

   • Case 1: $f(x) = 2$

   $2e^{x} - 5 = 2 \implies 2e^{x} = 7 \implies e^{x} = \frac{7}{2} \implies x = \ln\left(\frac{7}{2}\right)$

   • Case 2: $-f(x) = 2$

   $2e^{x} - 5 = -2 \implies 2e^{x} = 3 \implies e^{x} = \frac{3}{2} \implies x = \ln\left(\frac{3}{2}\right)$

The exact solutions are therefore $x = \ln\left(\frac{7}{2}\right)$ and $x = \ln\left(\frac{3}{2}\right)$.