The functions f and g are defined by $f: x \mapsto e^x + 2, \; x \in \mathbb{R}$ g: x \mapsto \ln x, \; x > 0$ (a) State the range of f - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 5

To find $g(f(x))$, we substitute $f(x)$ into $g(x)$:

$g(f(x)) = g(e^x + 2) = \ln(e^x + 2)$

This is the answer in its simplest form.

Start by solving the equation:

$f(2x + 3) = e^{2x + 3} + 2 = 6$

Subtract 2 from both sides:

$e^{2x + 3} = 4$

Taking the natural logarithm of both sides gives:

$2x + 3 = \ln 4$

Now, isolate x:

$2x = \ln 4 - 3$

$x = \frac{\ln 4 - 3}{2}$

To find the inverse function $f^{-1}(x)$, we start with:

$y = e^x + 2$

Now, switch the variables x and y and solve for y:

$x - 2 = e^y$

Taking the natural logarithm on both sides:

$y = \ln(x - 2)$

Thus, the inverse function is:

$f^{-1}(x) = \ln(x - 2)$

The domain of $f^{-1}$ is where $x - 2 > 0$, which leads to:

$\text{Domain of } f^{-1}: (2, \infty)$

To sketch the graphs of $y = f(x)$ and $y = f^{-1}(x)$:

• The graph of $y = f(x)$ starts from (0, 3) and increases to infinity. It does not cross the x-axis as $f(x) > 2$ for all x.
• The graph of $y = f^{-1}(x)$ will start from (2, 0) and increases, crossing the y-axis at (0, -\ln 2).
• The curves will cross at the point $(3, 3)$, where $f(1) = 3$. Therefore, points where they cross the axes include:
  • f(x): (0, 3), (x, 2) does not exist on f(x)
  • f^{-1}(x): (2, 0), (0, -\ln 2) where it approaches the asymptote as $x$ tends to 2.