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 (fg(x)):

1. We know (g(x) = \ln x) for (x > 0).
2. Therefore, (fg(x) = f(g(x)) = f(\ln x) = e^{\ln x} + 2 = x + 2).

So, (fg(x) = x + 2).

To solve for (x):

1. Set the equation: (f(2x + 3) = 6).
2. This gives us: (e^{(2x + 3)} + 2 = 6).
3. Subtract 2 from both sides: (e^{(2x + 3)} = 4).
4. Now, apply the natural logarithm: (2x + 3 = \ln 4).
5. Rearranging gives: (2x = \ln 4 - 3), thus (x = \frac{\ln 4 - 3}{2}).

To find the inverse function:

1. Start with (y = e^{x} + 2).
2. Rearranging gives (e^{x} = y - 2).
3. Now, take the natural log of both sides: (x = \ln(y - 2)).
4. Therefore, (f^{-1}(y) = \ln(y - 2)).
5. The domain of (f^{-1}) is (y > 2).

For sketching the curves:

1. The curve (y = f(x) = e^{x} + 2) will be shaped such that it crosses the y-axis at (y = 3) (when (x = 0)).
2. The curve (y = f^{-1}(x) = \ln(x - 2)) will cross the x-axis at (x = 2) because (f^{-1}(2) = 0).
3. Both curves are symmetric about the line (y = x).

Notable points where they cross the axes:

• (f(0) = 3)
• (f^{-1}(2) = 0)
• The curves will not cross the axes at other points!